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CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 6 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Tangents And Its Equations
Quiz 6
The points of contact of the vertical tangents $$x= 2-3\sin \theta $$, $$y= 3+2\cos \theta $$ are
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$$\left ( 2,5 \right ),\left ( 2,1 \right )$$
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$$\left ( -1,3 \right ),\left ( 5,3 \right )$$
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$$\left ( 2,5 \right ),\left ( 5,3 \right )$$
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$$\left ( -1,3 \right ),\left ( 2,1 \right )$$
Explanation
For the tangents to be vertical,
$$\dfrac{dy}{dx}=\infty$$
Or
$$\dfrac{dx}{dy}=0$$
Or
$$\dfrac{dx}{d\theta}=0$$
Or
$$-3\cos\theta=0$$
Or
$$\theta=\dfrac{2n-1}{2}\pi$$
Hence
$$\theta=\dfrac{\pi}{2},\dfrac{3\pi}{2}$$.
Now
$$x_{\tfrac{\pi}{2}}$$
$$=-1$$
$$y_{\tfrac{\pi}{2}}$$
$$=3$$.
Similarly
$$x_{\tfrac{3\pi}{2}}=5$$
$$y_{\tfrac{3\pi}{2}}=3$$.
Hence the points are
$$(-1,3)$$ and $$(3,5)$$.
Tangent is drawn to ellipse $$\displaystyle \frac{x^{2}}{27}+y^{2}=1$$ at $$\left ( 3\sqrt{3}\cos \theta ,\sin \theta \right )$$ (where $$\theta \in \left ( 0,\dfrac{\pi}2 \right )).$$ Then the value of $$\theta$$ such that sum of intercepts on axes made by this tangent is least is
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$$\dfrac {\pi}3$$
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$$\dfrac {\pi}6$$
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$$\dfrac {\pi}8$$
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$$\dfrac {\pi}4$$
Explanation
Let $$x=3\sqrt{3}\cos\theta$$
And
$$y=\sin\theta$$
Hence, $$\dfrac{dy}{dx}$$
$$=-\dfrac{\cos\theta}{3\sqrt{3}\sin\theta}$$ ...(i)
Hence equation of the tangent will be
$$y-\sin\theta=-\dfrac{\cos\theta}{3\sqrt{3}\sin\theta}(x-3\sqrt{3}.\cos\theta)$$
$$3\sqrt{3}\sin\theta.y-3\sqrt{3}\sin^{2}\theta=-x\cos\theta+3\sqrt{3}\cos^{2}\theta$$
$$3\sqrt{3}\sin\theta.y+x\cos\theta=3\sqrt{3}$$.
Hence the intercepts are
$$cosec\theta$$ and $$3\sqrt{3}.\sec\theta$$.
Hence, $$k=cosec\theta+3\sqrt{3}.\sec\theta$$
Now $$\dfrac{dk}{d\theta}$$ $$=-cosec\theta.\cot\theta+3\sqrt{3}.\sec\theta.\tan\theta$$ $$=0$$
$$3\sqrt{3}.\sec\theta.\tan\theta=cosec\theta.\cot\theta$$
$$3\sqrt{3}.\tan^{2}\theta.\dfrac{1}{\cos\theta}=\dfrac{1}{\sin\theta}$$
$$3\sqrt{3}.\tan^{2}\theta=\dfrac{\cos\theta}{\sin\theta}$$
$$3\sqrt{3}.\tan^{3}\theta=1$$
$$\tan^{3}\theta=\dfrac{1}{3\sqrt{3}}$$
$$\tan\theta=\dfrac{1}{\sqrt{3}}$$
$$\theta=\tan^{-1}(\dfrac{1}{\sqrt{3}})$$
$$\theta=30^{0}$$ $$=\dfrac{\pi}{6}$$.
The tangent to the curve $$\displaystyle 3xy^{2}-2x^{2}y=1$$ at $$(1, 1)$$ meets the curve again at the point
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$$\displaystyle \left ( \frac{16}{5},\frac{1}{20} \right )$$
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$$\displaystyle \left (- \frac{16}{5},-\frac{1}{20} \right )$$
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$$\displaystyle \left ( \frac{1}{20},\frac{16}{5} \right )$$
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$$\displaystyle \left ( -\frac{1}{20},\frac{16}{5} \right )$$
Explanation
Given, $$3xy^{2}-2x^{2}y=1$$
Differentiating with respect to x, gives us
$$3y^{2}+6xyy'-4xy-2x^{2}y'=0$$
At (1,1)
$$3+6y'-4-2y'=0$$
$$4y'=1$$
$$y'=\dfrac{1}{4}$$
Hence the equation of the tangent is
$$\dfrac{y-1}{x-1}=\dfrac{1}{4}$$
$$4y-4=x-1$$
$$x-4y=-3$$
$$x=4y-3$$
Substituting in the equation of the curve, give us
$$3(4y-3)y^{2}-2(4y-3)^{2}y=1$$
$$(4y-3)y[3y-2(4y-3)]=1$$
$$(4y-3)y(6-5y)=1$$
$$y(4y-3)(5y-6)=-1$$
$$y(20y^{2}-24y-15y+18)=-1$$
$$y(20y^{2}-39y+18)=-1$$
$$20y^{3}-39y^{2}+18y+1=0$$
$$y=1,\dfrac{-1}{20}$$
$$\Rightarrow x=1,\dfrac{-16}{5}$$
Hence the tangent again cuts the curve at $$(\dfrac{-16}{5},\dfrac{-1}{20})$$
The angle at which the curve $$y=ke^{kx}$$ intersects the $$y$$ -axis is
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$$\tan ^{-1}(k^{2})$$
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$$\cot ^{-1}(k^{2})$$
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$$\sin ^{-1}\left ( 1/\sqrt{1+k^{4}} \right )$$
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$$
\sec ^{-1}\left ( 1/\sqrt{1+k^{4}} \right )$$
Explanation
$$\displaystyle \dfrac{dy}{dx}=k^{2}e^{kx}$$.
The curve intersects $$y$$-axis at $$\left ( 0,k \right )$$
So, $$\displaystyle \dfrac{dy}{dx}|_{\left ( 0,k \right )}=k^{2}$$.
If $$\theta $$ is the angle at which the given
curve intersects the $$y$$-axis then $$\displaystyle \tan \left ( \pi /2-\theta \right )=\dfrac{k^{2}-0}{1+0.k^{2}}=k^{2}$$.
Hence $$\theta =\cot ^{-1}k^{2}$$
The lines tangent to the curves $$\displaystyle y^{3}-x^{2}y+5y-2x=0$$ and $$\displaystyle x^{4}-x^{3}y^{2}+5x+2y=0$$ at the origin intersect at an angle $$\displaystyle \theta $$ equal to
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$$\displaystyle \frac{\pi }{6}$$
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$$\displaystyle \frac{\pi }{4}$$
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$$\displaystyle \frac{\pi }{3}$$
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$$\displaystyle \frac{\pi }{2}$$
Let $$\displaystyle f\left ( x \right )=x^{3}+ax+b$$ with $$\displaystyle a\neq b$$ and suppose the tangent lines to the graph of $$f$$ at $$x = a$$ and $$x = b$$ have the same gradient Then the value of $$f (1)$$ is equal to
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$$0$$
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$$1$$
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$$\displaystyle -\frac{1}{3}$$
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$$\displaystyle \frac{2}{3}$$
Explanation
$$f(x)=x^3+ax+b$$
$$f'(x)=3x^2+a$$
Given gradient at $$x=a$$ and at $$x=b$$ are same
$$\Rightarrow 3a^2 +a=3b^2+a\Rightarrow b^2=a^2$$
But given $$a\neq b$$
$$\Rightarrow a+b=0 ..(1)$$
Hence $$ f(1)=1+a+b=1$$ using (1)
A curve with equation of the form $$\displaystyle y=ax^{4}+bx^{3}+cx+d$$ has zero gradient at the point (0, 1) and also touches the x-axis at the point (-1, 0) then the values of x for which the curve has a negative gradient are
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x > -1
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x < 1
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x < -1
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$$\displaystyle -1\leq \times \leq 1$$
Explanation
Given, $$\displaystyle y=ax^{4}+bx^{3}+cx+d$$
$$\Rightarrow y'=4ax^3+3bx^2+c$$
Using given conditions,
$$y(0)=1\Rightarrow d=1$$
$$y'(0)=0\Rightarrow c=0$$
$$y(-1)=0\Rightarrow a-b=-1 ..(1)$$
and $$y'(-1)=0\Rightarrow 4a-3b=0 ..(2)$$
Solving equation (1) and (2) we get, $$a=3,b=4$$
Hence the polynomial is,
$$y=3x^4+4x^3+1$$
$$y'=12x^2(1+x)$$
Now for negative gradient
$$y' < 0\Rightarrow 12x^2(1+x)< 0$$
$$\Rightarrow x< -1$$
For the curve represented parametrically by the equations $$\displaystyle x=2\ln\cot t+1$$ and $$\displaystyle y=\tan t+\cot t$$
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normal at
$$t = \displaystyle \dfrac {\pi }4$$
is parallel to y=axis
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tangent at $$t = \displaystyle \dfrac {\pi }4$$ is parallel to x-axis
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tangent at
$$t = \displaystyle \dfrac {\pi }4$$
is parallel to the line $$y = x$$
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normal at
$$t = \displaystyle \dfrac {\pi }4$$
is parallel to the line $$y = x$$
Explanation
Given, $$x= 2 \ln \cot t+1$$ and $$y=\tan t+\cot t$$
$$\therefore \displaystyle \frac{dx}{dt}=\frac{2\left ( -\mathrm{cosec} ^{2t} \right )}{\cot t}$$
at $$\displaystyle t = \frac{\pi }{4},\frac{dx}{dt}=-4$$
and $$\displaystyle \frac{dy}{dt}=\sec ^{2}t-\mathrm{cosec} ^{2}t$$
$$\displaystyle at t =\frac{\pi }{4}\:\:\;\frac{dy}{dt}=0$$
$$\displaystyle \frac{dy}{dx}=0$$ for tangent & hence it is parallel to x-axis & its normal is parallel to y axis
If a variable tangent to the curve $$\displaystyle x^{2}y=c^{3}$$ makes intercepts $$a, b$$ on $$x$$ and $$y$$ axis respectively then the value of $$\displaystyle x^{2}$$ is
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$$\displaystyle 27c^{3}$$
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$$\displaystyle \frac4{27}c^{3}$$
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$$\displaystyle \frac{27}{4}c^{3}$$
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$$\displaystyle \frac4{9}c^{3}$$
Explanation
Let any point on the curve is $$\displaystyle \left ( \frac{c}{t},ct^{2} \right )$$
$$\displaystyle y=\frac{c^{3}}{x^{2}}\Rightarrow \frac{dy}{dx}=-\frac{2c^{3}}{x^{3}}=-\frac{2c^{3}}{\dfrac {c^{3}}{t^{3}}}$$
$$\displaystyle \frac{dy}{dx}=-2t^{3}$$
Equation of tangent is
$$\displaystyle y-ct^{2}=-2t^{3}\left ( x-\frac{c}{t} \right )$$
for x intercept
$$\displaystyle 0-ct^{2}=-2t^{3}\left ( a-\frac{c}{t} \right )\Rightarrow \frac{c}{2t}=a-\frac{c}{t}$$
$$\displaystyle a=\frac{3}{2}\frac{c}{t}$$
for y intercept
$$\displaystyle b-ct^{2}=-2t^{3}\left ( 0-\frac{c}{t} \right )$$
$$\displaystyle \Rightarrow b-ct^{2}=2t^{2}c \:\:\Rightarrow b=3ct^{2}$$
$$\displaystyle a^{2}b=\frac{9}{4},\frac{c^{2}}{t^{2}}\times 3ct^{2}=\frac{27c^{3}}{4}$$
The coordinates of the point(s) on the graph of the function $$\displaystyle f(x)=\frac{x^{3}}3{-\frac{5x^{2}}{2}}+7x-4$$ where the tangent drawn cut off intercepts from the coordinate axes which are equal in magnitude but opposite in sign is
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$$(2,\dfrac 83)$$
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$$(3, \dfrac 72)$$
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$$(1,\dfrac 56)$$
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none
Explanation
Hence $$\dfrac{dy}{dx}=1$$ at that point.
Hence
$$x^{2}-5x+7=1$$
Or
$$x^{2}-5x+6=0$$
Or
$$(x-2)(x-3)=0$$
Or
$$x=2$$ and $$x=3$$
$$f(2)=\dfrac{8}{3}$$
And
$$f(3)=\dfrac{7}{2}$$
Hence the points are $$(2,\dfrac{8}{3})$$ and $$(3,\dfrac{7}{2})$$.
The number of values of c such that the straight line $$3x + 4y = c$$ touches the curve $$\displaystyle \frac{x^{4}}{2}=x+y$$ is
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$$0$$
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$$1$$
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$$2$$
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$$4$$
Explanation
$$3x+4y=c$$
Or
$$3+4y'=0$$
Or
$$y'=\dfrac{-3}{4}$$.
This is the slope of the tangent.
Now
Differentiating the equation of the curve with respect to $$x$$, we get
$$2x^{3}=1+y'$$
Now
$$y'=\dfrac{-3}{4}$$
Hence
$$2x^{3}=\dfrac{1}{4}$$
Or
$$x^{3}=\dfrac{1}{8}$$
Hence
$$x=0.5$$
Substituting in the equation of the curve we get
$$\dfrac{1}{32}=\dfrac{1}{2}+y$$
Or
$$y=\dfrac{-15}{32}$$.
Hence substituting the above points on the equation of the line, we get
$$\dfrac{3}{2}+\dfrac{-15}{8}=c$$
Or
$$c=\dfrac{-3}{8}$$
Hence there is only $$1$$ value of $$c$$.
Find all the tangents to the curve $$\displaystyle y=\cos \left ( x+y \right ),-2\pi \leq \times \leq 2\pi $$ that are parallel to the line $$x + 2y = 0$$
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$$\displaystyle x + 2\: y=\dfrac {\pi }2$$ & $$\displaystyle x +2y=-\dfrac {3\pi }2$$
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$$\displaystyle x + 2\: y=\dfrac {\pi }2$$ & $$\displaystyle x +2y=-\dfrac {\pi }2$$
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$$\displaystyle x + 2\: y=-\dfrac {3\pi }2$$ & $$\displaystyle x +2y=-\dfrac {3\pi }2$$
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$$\displaystyle x + 2\: y=\dfrac {3\pi }2$$ & $$\displaystyle x +2y=-\dfrac {3\pi }2$$
Explanation
Given equation is $$x+2y=0$$
Slope $$\displaystyle = - \frac{1}{2}$$
$$\displaystyle = - \frac{1}{2}=-sin(x+y)(1+y')$$
$$\displaystyle \Rightarrow -\frac{1}{2}=-\sin \left ( x+y \right )\left ( 1-\frac{1}{2} \right )$$
$$\displaystyle \Rightarrow \sin \left ( x+y \right )=1$$
$$\displaystyle \Rightarrow \cos \left ( x+y \right )=0\Rightarrow y=0\cos x=0$$
$$\displaystyle \therefore x=\frac{\pi }{2},\frac{3\pi }{2},-\frac{\pi }{2}.-\frac{3\pi }{2}$$
$$\displaystyle \sin x=1$$ is possible for x = $$\displaystyle\frac{\pi }{2}$$ or $$\displaystyle -\frac{3\pi }{2}$$
Equation are : $$\displaystyle y-0=-\frac{1}{2}\left ( x-\frac{\pi }{2} \right )$$ and $$\displaystyle y-0=-\frac{1}{2}\left ( x+\frac{3\pi }{2} \right )$$
The angle at which the curve $$\displaystyle y=ke^{kx}$$ intersects the y - axis is
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$$\displaystyle \tan ^{-1}k^{2}$$
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$$\displaystyle \cot ^{-1}\left ( k^{2} \right )$$
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$$\displaystyle \sin ^{-1\left ( \dfrac{1}{\sqrt{1+k^{4}}} \right )}$$
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$$\displaystyle \sec {-1\left ( \sqrt{1+k^{4}} \right )}$$
Explanation
$$\displaystyle \dfrac{dy}{dx}=K^{2}e^{kx}$$
$$\displaystyle \dfrac{dy}{dx}|_{x=0}=K^{2}=\tan \theta $$
(where $$\displaystyle \phi $$ is angle made by x-axis)
Let $$\displaystyle \phi $$ be the gngle made by y-axis
$$\displaystyle \tan \theta =\tan \left ( \dfrac{3\pi }{2}-\phi \right )=\cot \phi $$
$$\displaystyle \cot \phi =K^{2}$$
$$\displaystyle \phi =\cot ^{-1}\left ( K^{2} \right )$$
$$\displaystyle \Rightarrow \phi =\sin ^{-1}\left ( \dfrac{1}{\sqrt{1+K^{4}}} \right )$$
The abscissa of the point on the curve $$\displaystyle \sqrt{xy}=a+x$$ the tangent at which cuts off equal intercepts from the co-ordinate axes is (a > 0)
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$$\displaystyle \dfrac{a}{\sqrt{2}}$$
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$$\displaystyle- \dfrac{a}{\sqrt{2}}$$
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$$\displaystyle a\sqrt{2}$$
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$$\displaystyle -a\sqrt{2}$$
Explanation
$$\displaystyle y=\dfrac{\left (a+x \right )^{2}}{x}\Rightarrow y=\dfrac{a^{2}}{x}+2a+x$$
$$\displaystyle \dfrac{dy}{dx}=\dfrac{-a^{2}}{x^{2}}+1=-1$$ ( for equal intercepts)
$$\displaystyle x^{2}=\dfrac{a^{2}}{2}\Rightarrow x=\pm \dfrac{a}{\sqrt{2}}$$
Consider the curve $$\displaystyle f(x)=x^{1/3}$$ then
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the equation of tangent at (0, 0) is x = 0
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the equation of normal at (0, 0) is y = 0
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normal to the curve does not exist at (0, 0)
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f(x) and its inverse meet at exactly 3 points
Explanation
$$f(x)={x}^{1/3}\Rightarrow\displaystyle f'\left ( x \right )=\frac{1}{3x^{2/3}}$$
$$\displaystyle f'\left ( 0 \right )\rightarrow \infty$$ tangent is vertical at $$x = 0$$
Equation of tangent at $$(0,0)$$ is $$x = 0$$
Equation of normal is $$y = 0$$
$$\displaystyle f\left ( x \right )=f^{-1}\left ( x \right )$$
$$\implies \displaystyle x^{\frac{1}{3}}=x^{3}\implies x^{9}=x$$
$$\displaystyle \implies x = 0, 1, -1$$
Equation of the line through the point $$(1/2,2)$$ and tangent to the parabola $$\displaystyle y=\frac{-x^{2}}{2}+2$$ and secant to the curve $$\displaystyle y=\sqrt{4-x^{2}}$$ is
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$$2x + 2y - 5 = 0$$
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$$2x + 2y - 9 = 0$$
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$$y - 2 = 0$$
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none
Explanation
Equation of tangent is $$\displaystyle y-2=m\left ( x-\dfrac{1}{2} \right )$$
$$\displaystyle y=mx+2-\dfrac{m}{2}$$
Put it in the parabolas $$\displaystyle mx+2-\dfrac{m}{2}=-\dfrac{x^{2}}{2}+2$$
$$\displaystyle \dfrac{x^{2}}{2}+mx-\dfrac{m}{2}=0$$
since D = 0
$$\displaystyle \Rightarrow m^{2}+m=0$$
m = 0, -1 Two tangents are there
(i) y = 2
(ii)$$\displaystyle y=-x+2+\dfrac{1}{2}$$
$$\displaystyle \Rightarrow y=-x+\dfrac{5}{2}$$
$$\displaystyle y-=-x+\dfrac{5}{2}$$
The line y = 2 is tangent but $$\displaystyle y=-x+\dfrac{5}{2}$$ is secant for the curve
Equation of a trangent to the curve $$\displaystyle y\cot x=y^{3}\tan x$$ at the point where the abscissa is $$\displaystyle \frac{\pi }{4}$$ is
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$$\displaystyle 4x+2y=\pi +2$$
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$$\displaystyle 4x-2y=\pi +2$$
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x = 0
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y = 0
Explanation
Put $$x=\cfrac{\pi}{4}$$ in the given equation,
$$y=y^3\Rightarrow y(y^2-1)=0\Rightarrow y=-1,0,1$$
So the points are $$(\cfrac{\pi}{4}, -1),(\cfrac{\pi}{4}, 0)$$ and $$(\cfrac{\pi}{4}, 1)$$
Now given equation may be written as, $$y=\tan^2x.y^3$$
Differentiating w.r.t $$x$$
$$\cfrac{dy}{dx}=\tan^2x.3y^2\cfrac{dy}{dx}+y^3.2\tan x.\sec^2 x$$
$$\Rightarrow \cfrac{dy}{dx}=\cfrac{2y^3\sec^2 x.\tan x}{1-3\tan^2x.y^2}$$
Thus slope of tangents are,
$$m_1=\left(\cfrac{dy}{dx}\right)_{(\cfrac{\pi}{4}, -1)}=\cfrac{2(-1).2.1}{1-3.1.1}=2$$
$$m_2=\left(\cfrac{dy}{dx}\right)_{(\cfrac{\pi}{4},0)}=\cfrac{2(0).2.1}{1-3.1.0}=0$$
$$m_3=\left(\cfrac{dy}{dx}\right)_{(\cfrac{\pi}{4}, 1)}=\cfrac{2(1).2.1}{1-3.1.1}=-2$$
Hence equation of tangents are,
$$(y+1)=2(x-\cfrac{\pi}{4})\Rightarrow 4x-2y=\pi +2$$
$$ (y-0)=0(x-\cfrac{\pi}{4})\Rightarrow y=0$$
$$(y-1)=2(x-\cfrac{\pi}{4})\Rightarrow 4x+2y=\pi +2$$
If the curve $$\displaystyle { \left( \frac { x }{ a } \right) }^{ n }+{ \left( \frac { y }{ b } \right) }^{ n }=2$$ touches the straight line $$\displaystyle \frac { x }{ a } +\frac { y }{ b } =2$$, then find the value of $$n$$.
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$$2$$
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$$3$$
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$$4$$
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any real number
Explanation
Given $$\displaystyle { \left( \frac { x }{ a } \right) }^{ n }+{ \left( \frac { y }{ b } \right) }^{ n }=2$$
Differentiating both sides w.r.t $$x,$$ we get
$$\displaystyle \frac { n }{ a } { \left( \frac { x }{ a } \right) }^{ n-1 }+\frac { b }{ b } { \left( \frac { y }{ b } \right) }^{ n-1 }\times \frac { dy }{ dx } =0$$
$$\displaystyle \Rightarrow \frac { dy }{ dx } =-\frac { n }{ a } { \left( \frac { x }{ a } \right) }^{ n-1 }\times \frac { b }{ a } { \left( \frac { b }{ y } \right) }^{ n-1 }$$
$$\displaystyle \therefore \frac { dy }{ dx } $$ at $$\displaystyle \left( a,b \right) =\frac { b }{ a } $$
$$\therefore$$ Tangent is $$\displaystyle y-b=-\frac { b }{ a } \left( x-a \right) \Rightarrow bx+ay=2ab\Rightarrow \frac { x }{ a } +\frac { y }{ b } =2$$
for all values of $$n$$ $$(\because\displaystyle\frac{dy}{dx}$$ is independent of $$n)$$
Find the equation of the normal to the curve $$\displaystyle y = \left ( 1+x \right )^{y}+\sin ^{-1}\left ( \sin ^{2}x \right )$$ at $$x = 0$$
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$$x + y - 1 = 0$$
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$$x + y - 2 = 0$$
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$$x + y - 3 = 0$$
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$$x + y - 4 = 0$$
Explanation
Clearly at $$x=0 , y=1$$
Thus curve is, $$\displaystyle y = \left ( 1+x \right )^{y}+\sin ^{-1}\left ( \sin ^{2}x \right )$$
Differentiating w.r.t $$x$$
$$\cfrac{dy}{dx}=(1+x)^y[\cfrac{y}{1+x}+\log(1+x).\cfrac{dy}{dx}]+\cfrac{2\sin x.\cos x}{\sqrt{1+\sin^4x}}$$
Putting $$x=0 , y=1$$, we get, $$\cfrac{dy}{dx}=1=m$$ (say)
Thus slope of the normal is $$=-\cfrac{1}{m}=-1$$
Hence require equation is given by,
$$(y-1)=-1(x-0)$$
$$\Rightarrow x+y=1$$
Find the equation of normal to the curve $$\displaystyle x^{2}=4y$$ passing through the point $$(1, 2)$$
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$$x + y = 3$$
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$$x-y=3$$
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$$2x-y=4$$
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$$2x-3y=1$$
Explanation
$$x^2=4y$$
$$\Rightarrow \cfrac{dy}{dx}=\cfrac{x}{2}=m$$ (say)
Thus slope of normal at $$(1,2)$$ is $$m'=-\cfrac{1}{m}=\dfrac{-2}{x}=\dfrac{-2}{2}=-1$$
Hence, equation of normal at $$(1,2)$$ to the parabola is,
$$(y-2)=-1(x-1)$$
$$\Rightarrow x+y=3$$
A function is defined parametrically by the equations
x= $$\displaystyle 2t+t^{2}\sin \frac{1}{t}$$ if $$\displaystyle t\neq 0$$;
$$ 0 $$, otherwise
and
y = $$\displaystyle \frac{1}{t}\sin t^{2}$$ if $$\displaystyle t\neq 0$$; $$ 0 $$, otherwise
Find the equation of the tangent and normal at the point for t = 0 if they exist
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Tangent ;$$ 2y - x = 0;$$ Normal: $$2x + y = 0$$
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Tangent ;$$ 3y - x = 0;$$ Normal: $$2x + y = 0$$
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Tangent ;$$ 2y - x = 0;$$ Normal: $$3x + y = 0$$
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Tangent ;$$ 3y - x = 0;$$ Normal: $$3x + y = 0$$
Explanation
Given, $$x= 2t+ t^2 \sin \dfrac {1}{t}$$ and $$y=\dfrac {1}{t} \sin t^2$$
At $$t = 0$$ the point is origin
$$\displaystyle \frac{dx}{dt}=\lim_{t\rightarrow 0}\frac{2t+t^{2}\sin \dfrac 1t-0}{t}=2$$
$$\displaystyle \frac{dy}{dt}=\lim_{t-0}\frac{\dfrac{1}{t}\sin t^{2}}{t}=1$$
$$\displaystyle \frac{dy}{dx}=\dfrac{1}{2}$$
equation of tangent is $$\displaystyle y-0=\frac{1}{2}\left ( x-0 \right )\Rightarrow 2y-x=0$$
equation of normal is $$y - 0 = -2(x - 0)\Rightarrow y+2x=0$$
The curve $$\displaystyle y=ax^{3}+bx^{2}+cx+5$$ touches the $$x$$ - axis at $$P(-2, 0)$$ and cuts the $$y$$-axis at a point $$Q$$, where its gradient is $$3$$. Find $$a, b, c$$.
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$$\displaystyle a=-\frac{1}{5}, b=1,c=3$$
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$$\displaystyle a=-\frac{1}{4}, b=-1,c=4$$
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$$\displaystyle a=-\frac{1}{4}, b=0,c=3$$
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$$\displaystyle a=-\frac{1}{3}, b=1,c=-3$$
Explanation
Given, $$y= ax^3+bx^2+cx+5$$
$$\therefore \displaystyle \frac{dy}{dx}= 3ax^{2}+2bx+c$$
Since the curve touches $$x$$-axis at $$\left ( -2,0 \right )$$ so
$$\displaystyle \frac{dy}{dx}|_{\left ( -2,0 \right )}= 0\Rightarrow 12a-4b+c= 0$$ .... $$\left ( i \right )$$
The curve cut the $$y$$-axis at $$\left ( 0,8 \right )$$, so
$$\displaystyle \frac{dy}{dx}|_{\left ( 0,8 \right )}= 3\Rightarrow c= 3$$
Also the curve passes through $$\left ( -2,0 \right )$$, so
$$0= -8a+4b-2c+8\Rightarrow -8a+4b-2= 0$$ ..... $$ \left ( ii \right )$$
Solving $$ \left ( i \right )$$ and $$ \left ( ii \right )$$ $$a = -\dfrac {1}{4}$$, $$b =0$$
The slope of the normal to the curve $$\displaystyle x=a\left ( \theta -\sin \theta \right ),\: \: y=a\left ( 1-\cos \theta \right )$$ at point $$\displaystyle \theta =\dfrac{\pi }2$$ is
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$$0$$
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$$1$$
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$$-1$$
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$$\dfrac 1{\displaystyle \sqrt{2}}$$
Explanation
The slope of normal $$\displaystyle x=a\left ( \theta -\sin \theta
\right );\: \: \: \: y=a\left ( 1-\cos \theta \right )\: \: \: at\: \:
\: \theta =\dfrac{\pi }{2}$$
$$\displaystyle \left ( \dfrac{dx}{d\theta } \right )_{\theta =\dfrac{\pi }{2}}=a\left ( 1-\cos \theta \right )=a $$
$$\displaystyle \left ( \dfrac{dy}{d\theta } \right )_{\theta =\dfrac{\pi }{2}}=a\sin \theta =a $$
$$\therefore \left ( \dfrac{dy}{dx} \right )=1$$
Therefore, slope of normal at the given point is, $$m= \left ( -\dfrac{dx}{dy} \right )=-1$$
Hence, option 'C' is correct.
The equation of the normal to the curve $$\displaystyle y^{2}=4ax $$ at point $$(a, 2a)$$ is
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0%
$$x - y + a = 0$$
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$$x + y - 3a = 0$$
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$$x + 2y + 4a = 0$$
0%
$$x + y + 4a = 0$$
Explanation
$$\displaystyle y^{2}=4ax $$
$$\displaystyle 2y\frac{dy}{dx}=4a\Rightarrow \cfrac{dy}{dx}=\cfrac{2a}{y} $$
$$\therefore \displaystyle
\left ( \frac{dy}{dx} \right )_{\left ( a,2a \right )}=\left (
\frac{2a}{y} \right )_{\left ( a,2a \right )}=1 =m $$ (say)
$$\displaystyle \therefore $$ Slope of normal at the given point is $$=-\cfrac{1}{m}=-1$$
Therefore, the equation of normal is, $$(y - 2a) = -1 (x - a)$$
$$\Rightarrow x + y = 3a$$
Hence, option 'B' is correct.
The tangent at a point $$P$$ of a curve meets the $$y-$$ axis at $$A$$ and the line parallel to $$y-$$ axis at $$A$$, and the line parallel to $$y-$$ axis through $$P$$ meets the $$x-$$ axis at $$B$$. If area of $$\Delta OAB$$ is constant ($$O$$ being the origin). Then the curve is
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0%
$$cx^2-xy+k=0$$
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$$x^{2}+y^{2}=cx$$
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$$3x^{2}+4y^{2}=k$$
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$$xy-x^{2}y^{2}+kx=0$$
A tangent to the hyperbola $$\displaystyle y=\frac { x+9 }{ x+5 } $$ passing though the origin is
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$$x+25y=0$$
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$$5x+y=0$$
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$$5x-y=0$$
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$$x-25y=0$$
Explanation
$$\displaystyle y=\frac { x+9 }{ x+5 } =1+\frac { 4 }{ x+5 } $$
$$\displaystyle \frac { dy }{ dx } $$ at $$\displaystyle \left( { x }_{ 1 },{ y }_{ 1 } \right) =-\frac { 4 }{ { \left( { x }_{ 1 }+5 \right) }^{ 2 } } $$
Equation of tangent is
$$\displaystyle y-{ y }_{ 1 }=-\frac { 4 }{ { \left( { x }_{ 1 }+5 \right) }^{ 2 } } \left( x-{ x }_{ 1 } \right) \Rightarrow y-1-\frac { 4 }{ { x }_{ 1 }+5 } =-\frac { 4 }{ { \left( { x }_{ 1 }+5 \right) }^{ 2 } } \left( x-{ x }_{ 1 } \right) $$
Since it passes through origin $$(0,0)$$
$$\displaystyle -1-\frac { 4 }{ { x }_{ 1 }+5 } =-\frac { 4 }{ { \left( { x }_{ 1 }+5 \right) }^{ 2 } } \Rightarrow { \left( { x }_{ 1 }+5 \right) }^{ 2 }+4\left( { x }_{ 1 }+5 \right) +4{ x }_{ 1 }=0$$
$$\Rightarrow { { x }_{ 1 } }^{ 2 }+18{ x }_{ 1 }+45=0\Rightarrow \left( { x }_{ 1 }+15 \right) \left( { x }_{ 1 }+3 \right) =0\Rightarrow { x }_{ 1 }=15$$ or $${ x }_{ 1 }=-3$$
So equation of tangent is $$\displaystyle y-1-\frac { 4 }{ \left( -15+5 \right) } =-\frac { 4 }{ { \left( -15+5 \right) }^{ 2 } } \left( x+15 \right) $$
$$\displaystyle \Rightarrow y-1+\frac { 2 }{ 5 } =-\frac { 1 }{ 25 } \left( x+15 \right) \Rightarrow y-\frac { 3 }{ 5 } =-\frac { x }{ 25 } -\frac { 3 }{ 5 } \Rightarrow x+25y=0$$
asymptotes of the graph
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$$\displaystyle x=\frac{3\pi }{2}$$
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$$\displaystyle x=-\frac{\pi }{2}$$
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$$\displaystyle x=\frac{\pi }{2}$$
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$$\displaystyle x=-\frac{3\pi }{2}$$
Explanation
$$f'(x)=\dfrac{\cos(x)}{1+\sin(x)}$$
Or
$$\dfrac{dy}{dx}=\dfrac{\cos x}{1+\sin x}$$
For asymptotes,
$$\dfrac{dy}{dx}=\infty$$
Or
$$dx=0$$
Or
$$\sin(x)=-1$$
Or
$$x=\dfrac{3\pi}{2},\dfrac{-\pi}{2}$$.
Let $$f$$ be a continuous, differentiable and bijective function. If the tangent to $$y=f\left( x \right) $$ at $$x=b$$, then there exists at least one $$c\in \left( a,b \right) $$ such that
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$$f'\left( c \right) =0$$
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$$f'\left( c \right) >0$$
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$$f'\left( c \right) <0$$
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none of these
Explanation
Since the same line is tangent at one point $$x=a$$ and normal at other point $$x=b$$
$$\Rightarrow$$ Tangent at $$x=b$$ will be perpendicular to tangent at $$x=a$$
$$\Rightarrow$$ Slope of tangent changes from positive to negative or negative to positive. Therefore, it takes the value zero somewhere. Thus, there exists a point $$c\in \left( a,b \right) $$ where $$f'\left( c \right) =0$$.
The slope of normal to the curve $$\displaystyle y^{2}=4ax$$ at a point $$\displaystyle \left ( at^{2},2at \right )$$ is
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$$\dfrac 1 t$$
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$$t$$
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$$-t$$
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$$-\dfrac 1 t$$
Explanation
$$y^2=4ax$$
$$\Rightarrow 2y\cfrac{dy}{dx}=4a\Rightarrow \cfrac{dy}{dx}=\cfrac{2a}{y}$$
Therefore, slope of normal to the given curve is, $$=\left(-\cfrac{dx}{dy}\right)_{(at^2,2at)}=-\cfrac{2at}{2a}=-t$$
Hence, option 'C' is correct.
The coordinates of the point P on the graph of the function $$\displaystyle y=e^-{\left | x \right |},$$ where area of triangle made by tangent and the coordinate axis has the greatest area, is
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$$\displaystyle \left ( 1,\frac{1}{e} \right )$$
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$$\displaystyle \left ( -1,\frac{1}{e} \right )$$
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$$\displaystyle \left ( e,e^{-e} \right )$$
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none
Explanation
The curve of f(x) is symmetrical about y axis.
Let the point of contact of the tangent be $$(a,e^{-a})$$
Therefore
$$\dfrac{dy}{dx}_{(a,e^{-a})}=-e^{-a}$$
Hence the equation of the tangent will be
$$y-e^{-a}=-e^{-a}(x-a)$$
$$y-e^{-a}=-e^{-a}x+ae^{-a}$$
$$y+e^{-a}x=e^{-a}(1+a)$$
$$\dfrac{y}{e^{-a}(1+a)}+\dfrac{x}{1+a}=1$$
Hence the area formed by the tangent and the coordinate axes will be
$$=\dfrac{(x-intercept)(y-intercept)}{2}$$
$$A=\dfrac{e^{-a}(1+a)^{2}}{2}$$
Differentiating with respect to a, we get
$$\dfrac{dA}{da}=\dfrac{1}{2}[2(1+a)e^{-a}-(1+a)^{2}e^{-a}]$$
$$=\dfrac{(1+a)e^{-a}}{2}[2-(1+a)]$$
$$=\dfrac{(1+a)e^{-a}}{2}(1-a)$$
$$=0$$
Hence
$$a=\pm1$$
Thus we get the points as
$$(1,e^{-1})$$ and $$(-1,e^{-1})$$.
On the ellipse, $$4x^2\, +\, 9y^2\, =\, 1$$, the points at which the tangents are parallel to the line $$8x = 9y$$ are
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$$\left ( \displaystyle \frac{2}{5},\,\frac{1}{5} \right )$$
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$$\left ( -\displaystyle \frac{2}{5},\,\frac{1}{5} \right )$$
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$$\left ( -\displaystyle \frac{2}{5},\,-\frac{1}{5} \right )$$
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$$\left ( \displaystyle \frac{2}{5},\,-\frac{1}{5} \right )$$
Explanation
We have $${ 4x }^{ 2 }+9{ y }^{ 2 }=1$$ ...(1)
Differentiating w.r.t x, we get
$$\displaystyle 8x+18y\frac { dy }{ dx } =0\Rightarrow \frac { dy }{ dx } =-\frac { 4x }{ 9y } $$
The tangent at point $$\left( x,y \right) $$ will be parallel to the line $$8x=9y$$ if
$$\displaystyle \frac { -4x }{ 9y } =\frac { 8 }{ 9 } \Rightarrow x=-2y$$
Substituting this in (1), we get
$$\displaystyle 4{ \left( -2y \right) }^{ 2 }+{ 9y }^{ 2 }=1\Rightarrow { 25y }^{ 2 }=1\Rightarrow y=\pm \frac { 1 }{ 5 } $$
Thus the point where the tangent are parallel to $$8x=9y$$ are
$$\displaystyle \left( \frac { -2 }{ 5 } ,\frac { 1 }{ 5 } \right) $$ and $$\displaystyle \left( \frac { 2 }{ 5 } ,\frac { -1 }{ 5 } \right) $$
The slope of the tangent to the curve $$xy + ax - by = 0$$ at the point $$(1, 1)$$ is $$2$$ then values of $$a$$ and $$b$$ are respectively -
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$$1, 2$$
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$$2, 1$$
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$$3, 5$$
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None of these
Explanation
Given curve is $$xy + ax - by = 0$$
Slope of tangent at $$(1, 1)$$ is $$\displaystyle x.\frac{dy}{dx}+y+a-b.\frac{dy}{dx}=0 $$
$$\displaystyle
\left ( \frac{dy}{dx} \right )_{\left ( 1,1 \right )}=-\left [
\frac{\left ( a+y \right )}{x-b} \right ]_{\left ( 1,1 \right
)}=-\frac{\left ( a+1 \right )}{1-b} $$
$$\displaystyle \therefore -\frac{\left ( a+1 \right )}{1-b}=2 $$
So, $$2b - a = 3 $$ ..........(1)
$$\displaystyle \because (1, 1)$$ lies on curve $$xy + ax - by = 0$$
$$\displaystyle \therefore a-b=-1 $$ .........(2)
From (1) and (2), we get
$$a = 1, b = 2$$
Hence, option 'A' is correct.
It $$\displaystyle x=t^{2}$$ and $$y = 2t$$ then equation of normal at $$t = 1$$ is -
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$$x + y + 3 = 0$$
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$$x + y + 1 = 0$$
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$$x + y - 1 = 0$$
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$$x + y - 3 = 0$$
Explanation
$$\displaystyle x=t^{2},\: \: y=2t $$
$$\Rightarrow \displaystyle \frac{dx}{dt}=2t\: \: \: \: \: \:, \frac{dy}{dt}=2 $$
$$\Rightarrow \displaystyle \frac{dy}{dx}=\frac{1}{t}$$
$$\therefore$$ Slope of normal is $$=\displaystyle \left ( \frac{-dx}{dy} \right )_{t=1}=-t=-1 $$
Therefore, equation of normal is, $$(y - 2) = -1(x - 1)$$
$$\Rightarrow x + y = 3$$
Hence, option 'D' is correct.
The equation of the tangent to the curve $$\displaystyle \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=\frac{2}{\sqrt{a}}$$ at point $$(a, a)$$ is
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$$\displaystyle \frac{a}{\sqrt{x}}+\frac{a}{\sqrt{y}}={2}\sqrt a$$
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$$x + y = 2a$$
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$$\displaystyle \sqrt{x}+\sqrt{y}=2\sqrt{a}$$
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None of these
Explanation
Given curve is, $$\displaystyle \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=\frac{2}{\sqrt{a}}$$
$$\Rightarrow x^{-1/2}+y^{-1/2}=2a^{-1/2}$$
Differentiating w.r.t $$x$$
$$\dfrac{-1}{2}x^{-3/2}-\dfrac{1}{2}y^{-3/2}\cfrac{dy}{dx}=0$$
$$\Rightarrow \cfrac{dy}{dx}=-\cfrac{y\sqrt{y}}{x\sqrt{x}}$$
Thus slope of tangent at $$(a,a)$$ is $$m=-1$$
The equation of required tangent is given by,
$$(y-a)=-1(x-a)\Rightarrow x+y=2a$$
Hence, option 'B' is correct.
Consider the curved mirror $$y=f(x)$$ passing through $$(0, 6)$$ having the property that all light rays emerging from origin, after getting reflected from the mirror becomes parallel to $$x-$$ axis, then the equation of curve is
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$$y^2=4(x-y)$$ or $$y^2=36(9+x)$$
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$$y^2=4(1-x)$$ or $$y^2=36(9-x)$$
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$$y^2=4(1+x)$$ or $$y^2=36(9-x)$$
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$$None\ of\ these$$
The point where the tangent line to the curve $$\displaystyle y=e^{2x}$$ at $$(0, 1)$$ meets $$x$$ - axis is -
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$$(1, 0)$$
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$$(-1, 0)$$
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$$\displaystyle \left ( -\dfrac12,0 \right )$$
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None of these
Explanation
$$\displaystyle y=e^{2x}$$
$$\displaystyle \frac{dy}{dx}=e^{2x}\times 2 $$
$$\therefore \displaystyle \left ( \frac{dy}{dx} \right )_{\left ( 0,1 \right )}=2 $$
Thus equation of tangent at point $$(0,1)$$ is, $$y - 1 = 2(x - 0)\Rightarrow y=2x+1$$
Now substitute $$y = 0$$ to get point of intersection of tangent with $$x-axis$$
$$\displaystyle x=-\frac{1}{2}$$
Therefore, required point is $$\displaystyle \left ( -\frac{1}{2},0 \right )$$
Hence, option 'C' is correct.
The slope of the tangents to the curve $$y = (x + 1) (x - 3)$$ at the points where it crosses x - axis are
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$$\displaystyle \pm 2 $$
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$$\displaystyle \pm 3 $$
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$$\displaystyle \pm 4$$
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None of these
Explanation
$$y = (x + 1)(x - 3) ..(1)$$
Substitute $$y = 0$$ to get the point of intersection of this curve with $$x-axis$$
$$0 = (x + 1) (x - 3)$$
$$\Rightarrow x = -1, 3$$
So the points are $$(-1, 0)$$ and $$(3, 0)$$
Now differentiating eq. (1), we get
$$\displaystyle \dfrac{dy}{dx}=\left ( x-3 \right )+\left ( x+1 \right )=2x-2 $$
$$\Rightarrow \displaystyle \left ( \dfrac{dy}{dx} \right )=2\left ( x-1 \right )$$
Thus slope at $$(-1,0)$$ is $$=\left (\dfrac{dy}{dx} \right )_{\left ( -1,0 \right )}=-4 $$
and at $$(3,0)$$ is $$=\displaystyle \left ( \dfrac{dy}{dx} \right )_{\left ( 3,0 \right )}=4 $$
Hence, option 'C' is correct.
The equation of tangent at the point $$\displaystyle \left ( at^{2},at^{3} \right )$$ on the curve $$\displaystyle ay^{2}=x^{3}$$ is
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$$\displaystyle 3tx-2y=at^{3}$$
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$$\displaystyle tx-3y=at^{3}$$
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$$\displaystyle 3tx+2y=at^{3}$$
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None of these
The equation of the normal to the curve $$ \left (\displaystyle x=at^{2} \right ), \left (y=2at \right )$$ at '$$t$$' point is -
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$$\displaystyle ty=x+at^{2}$$
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$$\displaystyle y+tx-2at-at^{3}=0$$
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$$\displaystyle y=tx-2at-at^{3}$$
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None of these
Explanation
Given $$\displaystyle x=at^{2},y=2at $$
$$\Rightarrow \cfrac{dx}{dt}=2at, \cfrac{dy}{dt}=2a$$
Thus slope of normal at point 't' is $$=-\cfrac{dx}{dy}=\dfrac{dx/dt}{dy/dt}=-t$$
Therefore, the equation of normal at $$'t'$$ is given by,
$$(y-2at)=-t(x-at^2)$$
$$\Rightarrow y+tx-2at-at^3=0$$
Hence, option 'B' is correct.
The equation of the tangent to the curve is $$\displaystyle y=2\sin x+\sin 2x$$ at the point $$\displaystyle x=\dfrac {\pi }3$$ is -
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$$\displaystyle 2y=\sqrt{3}$$
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$$\displaystyle 3y=\sqrt{2}$$
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$$\displaystyle 2y=3\sqrt{3}$$
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$$2y = 3$$
Explanation
$$\displaystyle y=2\sin x+\sin 2x $$
$$\displaystyle \frac{dy}{dx}=2\cos x+2\cos 2x $$
The slope of tangent at $$x=\cfrac{\pi}{3}$$ is, $$m=\displaystyle
\left ( \dfrac{dy}{dx} \right )_{x=\dfrac{\pi }{3}}=2\times
\dfrac{1}{2}+2\times \left ( -\dfrac{1}{2} \right )=1-1=0 $$
Now at $$\displaystyle x=\dfrac{\pi }{3};\: \: \: y=2\times \dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2} $$
$$\displaystyle y=\dfrac{\sqrt{3}}{2}\times 3 $$
So the point is $$\displaystyle \left ( \frac{\pi }{3},\frac{3\sqrt{3}}{2} \right )$$
Therefore, required tangent is,
$$\displaystyle \left ( y-\dfrac{3\sqrt{3}}{2} \right )=0$$
$$\Rightarrow y=\dfrac{3\sqrt{3}}{2}$$
$$\Rightarrow 2y=3\sqrt{3}$$
Hence, option 'C' is correct.
The equation of the tangent to the curve $$\displaystyle \sqrt{x}+\sqrt{y}=\sqrt{a}$$ at the point $$\displaystyle \left ( x_{1},y_{1} \right )$$ is -
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$$\displaystyle \frac{x}{\sqrt{x_{1}}}+\frac{y}{\sqrt{y_{1}}}=\frac{1}{\sqrt{a}}$$
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$$\displaystyle \frac{x}{\sqrt{x_{1}}}+\frac{y}{\sqrt{y_{1}}}=\sqrt{a}$$
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$$\displaystyle x\sqrt{x_{1}}+y\sqrt{y_{1}}=\sqrt{a}$$
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None of these
Explanation
$$\displaystyle \sqrt{x}+\sqrt{y}=\sqrt{a}$$ at point $$\displaystyle \left ( x_{1},y_{1} \right )$$
$$\displaystyle \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0 $$
$$\displaystyle
\frac{dy}{dx}=-\sqrt{\frac{y}{x}}\: \: \Rightarrow \: \: \left (
\frac{dy}{dx} \right )_{\left ( x_{1},y_{1} \right
)}=-\frac{\sqrt{y_{1}}}{\sqrt{x_{1}}}$$
Thus equation of tangent is given by, $$\displaystyle y-y_{1}=-\frac{\sqrt{y_{1}}}{\sqrt{x_{1}}}\left ( x-x_{1} \right )$$
$$\displaystyle y\sqrt{x_{1}}-y_{1}\sqrt{x_{1}}=-x\sqrt{y_{1}}+x_{1}\sqrt{y_{1}}$$
$$\displaystyle x\sqrt{y_{1}}+y\sqrt{x_{1}}=\sqrt{x_{1}}\sqrt{y_{1}}\left ( \sqrt{a} \right )$$
$$\displaystyle \frac{x}{\sqrt{x_{1}}}+\frac{y}{\sqrt{y_{1}}}=\sqrt{a}$$
Hence, option 'B' is correct.
The coordinates of the point on the curve $$\displaystyle y=x^{2}+3x+4$$ the tangent at which passes through the origin are -
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$$(-2, 2), (2, 14)$$
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$$(1, -1), (3, 4)$$
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$$(2, 14), (2, 2)$$
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$$(1, 2), (14, 3)$$
Explanation
Let the point is $$P(a,b)$$
Now the given curve is $$y=x^2+3x+4$$
Differentiating w.r.t $$x$$
$$\cfrac{dy}{dx}=2x+3$$
Thus slope of tangent at P is $$=\left(\cfrac{dy}{dx}\right)_{(a,b)}=2a+3$$
Hence equation of tangent at P is given by, $$(y-b)=(2a+3)(x-a)$$
But given that tangent passes through origin
$$\Rightarrow (0-b)=(2a+3)(0-a)\Rightarrow 2a^2+3a=b ..(1)$$
Also the point P lies on the given curve,
$$b=a^2+3a+4 ...(2)$$
Solving (1) and (2) we get the required point P coordinate
which are $$(-2,2) \mbox{or} (2,14)$$
Hence, option 'A' is correct.
At what point the tangent line to the curve $$\displaystyle y=\cos \left ( x+y \right ),\left ( -2\pi \leq x\leq 2\pi \right )$$ is parallel to $$x + 2y = 0$$
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$$\displaystyle \left ( \dfrac {\pi }2, 0 \right )$$
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$$\displaystyle \left ( -\dfrac {\pi }2, 0 \right )$$
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$$\displaystyle \left (\dfrac{ 3\pi }2, 0 \right )$$
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$$\displaystyle \left (-\dfrac{ 3\pi }2,\dfrac { \pi }2 \right )$$
Explanation
$$\displaystyle y=\cos \left ( x+y \right )$$ $$\displaystyle \left ( -2\pi \leq x\leq 2\pi \right )$$
$$\Rightarrow \displaystyle \frac{dy}{dx}=-\sin \left ( x+y \right )\left ( 1+\frac{dy}{dx} \right )$$
$$\Rightarrow \displaystyle \frac{dy}{dx}\left ( 1+\sin \left ( x+y \right ) \right )=-\sin \left ( x+y \right )$$
$$\Rightarrow \displaystyle \frac{dy}{dx}=-\frac{\sin \left ( x+y \right )}{1+\sin \left ( x+y \right )}$$
Given tangent is parallel to the line, $$x + 2y = 0$$
$$\Rightarrow \displaystyle \frac{dy}{dx}=-\frac{\sin \left ( x+y \right )}{1+\sin \left ( x+y \right )}=-\frac{1}{2}$$
$$\displaystyle 2\sin \left ( x+y \right )=1+\sin \left ( x+y \right )$$
$$\displaystyle \sin \left ( x+y \right )=1\Rightarrow \cos(x+y)=0\Rightarrow x+y=\cfrac{\pi}{2}$$
Also the point lies on the given curve,
$$y=0\Rightarrow x= \dfrac{\pi }{2}$$
Thus the point is, $$\displaystyle \left ( \frac{\pi }{2},0 \right )$$
Hence, option 'A' is correct.
The point at which the tangent to the curve $$\displaystyle y=x^{3}+5$$ is perpendicular to the line $$x + 3y = 2$$ are
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$$(6, 1), (-1, 4)$$
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$$(6, 1) (4, -1)$$
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$$(1, 6), (1, 4)$$
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$$(1, 6), (-1, 4)$$
Explanation
Let point be $$\displaystyle \left ( x_{1},y_{1} \right )$$
$$y = \displaystyle x^{3}+5$$
$$\Rightarrow \displaystyle \dfrac{dy}{dx}=3x^{2}$$
Now, the slope of tangent at this point is $$=\displaystyle \left ( \frac{dy}{dx} \right )_{x_{1}y_{1}}=3x_{1}^{2}$$
It is $$\displaystyle \perp $$ to $$x + 3y - 2 = 0$$
So $$\displaystyle 3x_{1}^{2}\times -\frac{1}{3}=-1\Rightarrow x_1=\pm 1 $$
Thus,
at $$\displaystyle x_{1}=1$$, $$\Rightarrow$$ $$\displaystyle y_{1}=6$$
and at $$\displaystyle x_{1}=-1$$ $$\Rightarrow$$ $$\displaystyle y_{1}=4$$
Thus, the points are $$(1, 6)$$ and $$(-1, 4)$$
Hence, option 'D' is correct.
A normal $$P(x,y)$$ on a curve meets the $$X-$$axis at $$Q$$ and $$N$$ is the ordinate at $$P$$.
If $$NQ=\dfrac {x(1+y^2)}{1+x^2} $$.
Then the equation of curves passing through $$(3,1)$$ is
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$$5(1+y^2)=(1+x^2)$$
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$$5(1+y^2)=5(1+x^2)$$
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$$5(1+x^2)=(1+y^2)$$
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None of these
At what point of the curve $$\displaystyle y=2x^{2}-x+1$$ tangent is parallel to $$y = 3x + 4$$
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$$(0, 1)$$
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$$(1, 2)$$
0%
$$(-1, 4)$$
0%
$$(2, 7)$$
Explanation
Let the point is $$P(a,b)$$
Now the given curve is $$y=2x^2-x+1$$
Differentiating w.r.t $$x$$
$$\cfrac{dy}{dx}=4x-1$$
Thus slope of tangent at P is $$=\left(\cfrac{dy}{dx}\right)_{(a,b)}=4a-1$$
But given the tangent is parallel to line $$y=3x+4$$
$$\Rightarrow 4a-1=3\Rightarrow a=1$$
Also the point P lies on the given curve,
$$b=2a^2-a+1=2$$
Therefore, the point P is $$(1,2)$$
Hence, option 'B' is correct.
Tangents are drawn from origin to the curve $$\displaystyle y=\sin x$$ then point of contect lies on -
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$$\displaystyle x^{2}=y^{2}$$
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$$\displaystyle x^{2}y^{2}=0$$
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$$\displaystyle x^{2}y^{2}=x^{2}-y^{2}$$
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None of these
Explanation
$$y=\sin x$$
$$\Rightarrow \cfrac{dy}{dx}=\cos x$$
Let the point of contact of tangent passing through origin to the given curve is $$P(h,k)$$
So equation of tangent to this curve passes through $$(0,0)$$ is,
$$y=(\cos h) x$$
For point of contact this line will pass through point P,
$$\Rightarrow k=h\cos h\Rightarrow \cos h=\cfrac{k}{h} ...(1)$$
Also point P lies on the given curve,
$$\sin h=k ..(1)$$
Now using $$(1)^2+(2)^2=1$$
$$\cfrac{k^2}{h^2}+k^2=1\Rightarrow k^2h^2=h^2-k^2$$
Thus locus of $$P(h,k)$$ is,
$$x^2y^2=x^2-y^2$$
Hence, option 'C' is correct.
A tangent to the curve $$\displaystyle y=x^{2}+3x$$ passes through a point $$(0, -9)$$ if it is drawn at the point -
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$$(-3, 0)$$
0%
$$(1, 4)$$
0%
$$(0, 0)$$
0%
$$(-4, 4)$$
Explanation
Let the point is $$\displaystyle P\left ( x_{1},y_{1} \right )$$ on the curve
So, $$\displaystyle \frac{dy}{dx}=2x+3 $$
$$\Rightarrow \displaystyle \left ( \frac{dy}{dx} \right )_{\left ( x_{1},y_{1} \right )}=2x_{1}+3 $$
Therefore, the tangent is $$\displaystyle \frac{y-y_{1}}{x-x_{1}}=2x_{1}+3 $$
$$\displaystyle \left ( y-y_{1} \right )=\left ( x-x_{1} \right )\left ( 2x_{1}+3 \right )$$
$$\displaystyle y-y_{1}=2xx_{1}+3x-2x_{1}^{2}-3x_{1} $$
$$\displaystyle x\left ( 2x_{1}+3 \right )-y+y_{1}-2x_{1}^{2}-3x_{1}=0$$
It passes $$(0, -9)$$ then
$$\displaystyle 0\times \left ( 2x_{1}+3 \right )+9+y_{1}-2x_{1}^{2}-3x_{1}=0 $$
$$\displaystyle \left ( x_{1},y_{1} \right )$$ also lies on the curve
$$=\displaystyle y_{1}-2x_{1}^{2}-3x_{1}+9=0 $$ ..........(1)
So $$\displaystyle y_{1}=x_{1}^{2}+3x_{1} $$ ..........(2)
From eq (1) and (2), we get
$$\displaystyle y_{1}-2x_{1}^{2}-3x_{1}+9-y_{1}+x_{1}^{2}+3x_{1}=0 $$
$$\displaystyle -x_{1}^{2}+9=0 $$
$$\displaystyle x_{1}^{2}=9 $$
$$\displaystyle x_{1}=\pm 3 $$
when $$\displaystyle x_{1}= 3 $$ $$\Rightarrow$$ $$\displaystyle y_{1}= 9+9=18 $$
$$\displaystyle x_{1}= -3 $$ $$\Rightarrow$$ $$\displaystyle y_{1}= 9-9=0 $$
Thus the required points are $$(-3, 0)$$ and $$(3, 18)$$
Hence, option 'A' is correct.
The equation of the normal to the curve $$\displaystyle y^{2}=x^{3}$$ at the point whose abscissa is $$8$$ is -
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$$\displaystyle x\pm \sqrt{2}y=104$$
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$$\displaystyle x\pm 3\sqrt{2}y=104$$
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$$\displaystyle 3\sqrt{2}x\pm y=104$$
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None of these
Explanation
Solving the curve at $$x=8=2^3$$
$$\displaystyle y^{2}=x^{3}=2^{9}$$
$$\Rightarrow \displaystyle y=\pm 2^{4}\sqrt{2}=\pm 16\sqrt{2}$$
So the points are $$\displaystyle \left ( 8,\pm 16\sqrt{2} \right )$$
Now the curve is, $$\displaystyle y^{2}=x^{3}$$
$$\Rightarrow \displaystyle 2y\times \frac{dy}{dx}=3x^{2}$$
$$\Rightarrow \displaystyle \left ( \frac{dy}{dx} \right )=\frac{3x^{2}}{2y}$$
$$\Rightarrow \displaystyle \left ( \frac{dy}{dx} \right )_{\left ( 8,\pm 16\sqrt{2} \right )}=\frac{3\times 64}{\pm 2\times 16\sqrt{2}}=\pm 3\sqrt{2}$$
Thus the slope of normal is $$=\displaystyle \left ( \frac{dy}{dx} \right )=\mp \frac{1}{3\sqrt{2}}$$
Therefore, equation of normal is,
$$\displaystyle \left ( y\mp 16\sqrt{2} \right )=\mp \frac{1}{3\sqrt{2}}\left ( x-8 \right )$$
$$\Rightarrow \displaystyle \mp 3\sqrt{2}y\pm 96=(x-8)$$
$$\Rightarrow \displaystyle x\pm 3\sqrt{2}y=104 $$
Hence, option 'B' is correct.
The normal to the curve $$\displaystyle \sqrt{x}+\sqrt{y}=\sqrt{a}$$ is perpendicular to $$x$$ axis at the point
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$$(0, a)$$
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$$(a, 0)$$
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$$(\dfrac a 4, \dfrac a 4)$$
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No where
Explanation
Let the point be $$P\displaystyle \left ( x_{1},y_{1} \right )$$
Now $$\displaystyle \sqrt{x}+\sqrt{y}=\sqrt{a}$$
Differentiating w.r.t $$x$$
$$\displaystyle \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\times \frac{dy}{dx}=0 $$
$$\displaystyle \frac{dy}{dx}=-\sqrt{\frac{y}{x}}$$
Thus, slope of normal at P is $$ =\displaystyle -( \frac{dx}{dy} )_{( x_{1},y_{1} )}=-\sqrt{\frac{x_{1}}{y_{1}}} $$
If normal $$\displaystyle \perp $$ to x axis then $$\displaystyle \frac{dx}{dy}=\frac{a}{0} $$
$$\displaystyle \therefore y_1 = a\Rightarrow x_1 = 0$$
Therefore, required point is $$(a, 0)$$
Hence, option 'B' is correct.
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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