is a normal to the circle $${x}^{2}+{y}^{2}=25$$

is a tangent to the circle $${x}^{2}+{y}^{2}=25$$

does not meet the circle $${x}^{2}+{y}^{2}=25$$

MCQ Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 9

Tangents are drawn from a point on the circle

$x^2+y^2=25$ to the ellipse

$9x^2+16y^2-144=0$ then the angle between the tangents is

0%
$\frac{\pi}{4}$
0%
$\frac{3\pi}{4}$
0%
$\frac{\pi}{2}$
0%
$\frac{2\pi}{3}$

Slope of the line

$\sqrt { { x }^{ 2 }+{ 4y }^{ 2 }-4xy+4 } +x-2y=1$ equals to

0%
$\dfrac { 1 }{ 2 }$
0%
$2$
0%
$-\dfrac { 1 }{ 2 }$
0%
$None\ of\ these$

Let f be a real-valued differentiable function on R (the set of all real numbers) such that

$f(1)=1$ . If the y-intercept of the tangent at any point P(x, y) on the curve

$y=f(x)$ is equal to the cube of the abscissa of P, then the value of

$f(-3)$ is equal to?

0%
$-3$
0%
$3$
0%
$9$
0%
$-9$

Explanation

$P\equiv (x,y)$

Let slope by

$\cfrac { dy }{ dx }$

So, equation of tangent will be

$(X-m)\cfrac { dy }{ dx } =(Y-y)$

for Y- intercept ,

$x=0$

So,

$Y=y-x\cfrac { dy }{ dx }$

Now according to question ,

${ x }^{ 3 }=y-x\cfrac { dy }{ dx } \\ \cfrac { dy }{ dx } -\cfrac { y }{ x } ={ x }^{ 2 }\\ \cfrac { xdy-ydx }{ { x }^{ 2 } } =-xdx\\ d(\cfrac { y }{ x } )=-xdx$

On integrating , we get

$\cfrac { y }{ x } =\cfrac { -{ x }^{ 2 } }{ 2 } +C$

Now,

$y(1)=1$

So, we get

$\cfrac { y }{ x } =\cfrac { -{ x }^{ 2 } }{ 2 } +\cfrac { 3 }{ 2 }$

So,

$y(-3)=f(-3)=9$

The inclination of the tangent at

$\theta = \dfrac {\pi}{3}$ on the curve

$x = a(\theta + \sin \theta), y = a(1 + \cos \theta)$ is

0%
$\dfrac {\pi}{3}$
0%
$\dfrac {\pi}{6}$
0%
$\dfrac {2\pi}{3}$
0%
$\dfrac {5\pi}{6}$

Explanation

For the curve ,

$\displaystyle \dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} = \dfrac{\dfrac{d(a(1+cos\theta))}{d\theta}}{\dfrac{d(a(\theta+sin\theta))}{d\theta}} = \dfrac{-asin\theta}{a+acos\theta} = \dfrac{-sin\theta}{1+cos\theta}$

So, If the inclination of tangent at

$\theta = \dfrac{\pi}{3}$ be

$y$ ,

Then,

$\tan(y) =$

$\dfrac{-sin(\dfrac{\pi}{3})}{1+cos\dfrac{\pi}{3}} = \dfrac{-1}{\sqrt{3}}$

Thus,

$y = \pi-\dfrac{\pi}{6}= \dfrac{5\pi}{6}$

Equation of a normal to the curve

$y=x\log{x}$ , parallel to

$2x-2y+3=0$ is

0%
$x+y=3e^{-2}$
0%
$x-y=3e^{-2}$
0%
$x-y=3e^{2}$
0%
$x+y=3e^{2}$

The greatest slope among the lines represented by the equation

$4x^2 - y^2 + 2y - 1 = 0$ is -

0%
$-3$
0%
$-2$
0%
$2$
0%
$3$

Explanation

$4{x}^{2} - {y}^{2} + 2y - 1 = 0$

${y}^{2} - 2y + 1 = 4{x}^{2}$

${\left( y - 1 \right)}^{2} = {\left( 2x \right)}^{2}$

$y - 1 = \pm 2x$

$y = 2x + 1 \text{ or } y = -2x + 1$

Comparing the above equations with

$y = mx + c$ , we have

Slope

$\left( m \right) = 2, -2$

Hence the greatest slope will be

$2$ .

The ordinate of all points on the curve

$y=\dfrac{1}{2\sin^{2}x+3\cos^{2}x}$ where the tangent is horizontal, is

0%
Always equal to $\dfrac{1}{2}$
0%
Always equal to $\dfrac{1}{3}$
0%
$\dfrac{1}{2}$ or $\dfrac{1}{3}$ according as $n$ is an even or an odd integer.
0%
$\dfrac{1}{2}$ or $\dfrac{1}{3}$ according as $n$ is an even integer

Explanation

$y=\dfrac{1}{2sin^{2}x+3cos^{2}x}=\dfrac{1}{2+cos^{2}x}$

$tan\theta =\dfrac{dy}{dx}= \dfrac{-2sinx\, cos x}{(2+cos^{2}x)^{2}}=0$

$\therefore sin\, x=0$ or

$cos\, x= 0$

$sin\, x=0, y= \dfrac{1}{3}$ , if

$cos\, x=0, y= \dfrac{1}{2}$

1098464_1141711_ans_de42566bc3344c2b9b551bd59f7cb5c7.jpeg

Equation of a tangent to the curve

$y=\cos(x+y),\ 0\le x\le 2\pi$ that is parallel to the line

$x+2y=0$ is

0%
$x+2y=\pi/2$
0%
$x+2y=\pi/4$
0%
$x+2y=\pi$
0%
$x+y=\pi$

Explanation

Given curve y=cos (x+y)

we have find equation of tangent parallel to x+2y=0

slope of tangent is

$\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\dfrac{d}{dx} cos(x+y)=- \sin (x+y)\left(1+\dfrac{dy}{dx}\right)$

$\Rightarrow \dfrac{dy}{dx}(dx)(1+ \sin (x+y))=- \sin (x+y)$

Now slope x+2y=0 is

$1+2\dfrac{dy}{dx}=0 \Rightarrow \dfrac{dy}{dx}=\dfrac{-1}{2}$

Now

$+\dfrac{\sin (n+y)}{1+ \sin (x+y)}=+\dfrac{1}{2}$

$2 \sin (x+y)=1+ \sin (x+y)$

$\Rightarrow x+y=(2n+1)\dfrac{\pi }{2}$ for n= positive integer

$x=(2n+1)\dfrac{\pi }{2}-y$

$\therefore y=cos [(2n+1)\dfrac{\pi }{2}-y+y]= \cos \left[(2n+1)\dfrac{\pi }{2}\right]$

$\Rightarrow y=0$

Now

$2x\leq x\leq 2x$

putting n=0

$x=\dfrac{\pi }{2}$

n=-1

$x=-\dfrac{3\pi }{2}$

Thus i x=

$-\dfrac{3\pi }{2}$ &

$x=\dfrac{\pi }{2}$

$\therefore$ Points are

$\left(-\dfrac{3\pi }{2},0\right)$ and

$\left(\dfrac{\pi }{2},0\right)$

Now equation of tangent at

$\left(-\dfrac{3\pi }{2},0\right)$ and

$\left(\dfrac{\pi }{2},0\right)$ having solve

$-\dfrac{1}{2}$ is

$2x+4y+3\pi =0$

and

$2x+4y-\pi =0$

The curve given by

$x + y = {e^{xy}}$ has an tangents parallel to the y-axis at the point

0%
$(0,1)$
0%
$(1,0)$
0%
$(1,1)$
0%
$(0,0)$

Explanation

$\dfrac{dy}{dx}$ (differentiate the terms)

$1+\dfrac{dy}{dx}=e^{xy}\left(x\dfrac{dy}{dx}+y\right)$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{ye^{xy}-1}{1-xe^{xy}}$

Check by option

$\dfrac{dy}{dx}\left|_{(0, 1)}\right.=\dfrac{1-1}{1-0}=\dfrac{0}{1}=0$ Not coming infinity

$\displaystyle\dfrac{dy}{dx}\left|_{(1, 0)}\right.=\dfrac{-1}{0}=\infty$

This point satisfy

$\therefore$ option B

$(1, 0)$ .

1356400_1131715_ans_86b17a40d6924052a2042284677f440a.png

The equation of normal to the curve

$x^{3}+y^{3}=8xy$ at points where it is meet by the curve

$y^{2}=4x$ ,other then origin is

0%
$y=x$
0%
$y=-x+4$
0%
$y=2x$
0%
$y=-2x$

Tangents are drawn from origin to the curve

$y=\sin{x}+\cos{x}$ . Then their points of contact lie on the curve

0%
$\dfrac{1}{x^{2}}+\dfrac{2}{y^{2}}=1$
0%
$\dfrac{1}{x^{2}}-\dfrac{2}{y^{2}}=-1$
0%
$\dfrac{2}{x^{2}}+\dfrac{2}{y^{2}}=1$
0%
$\dfrac{2}{x^{2}}-\dfrac{2}{y^{2}}=1$

The slope of normal to the curve y= log (logx) at x = e is

0%
e
0%
-e
0%
$\frac{1}{e}$
0%
- $\frac{1}{e}$

Explanation

$\begin{array}{l} \frac { d }{ { dx } } \log \left( { \log x } \right) \\ Substitute\, u=\log x \\ =\dfrac { d }{ { du } } \log u\frac { d }{ { dx } } \log x \\ =\dfrac { 1 }{ u } \left( { \frac { 1 }{ x } } \right) \\ =\dfrac { 1 }{ { \log x } } \left( { \frac { 1 }{ x } } \right) \\ Substitute\, x=e \\ =\dfrac { 1 }{ { \log e } } \left( { \dfrac { 1 }{ e } } \right) \\ =\dfrac { 1 }{ e } \end{array}$

Number of possible tangents to the curve $y = \cos \left( {x + y} \right), - 3\pi \leqslant x \leqslant 3\pi$ , that are parallel to the line $x + 2y = 0$ , is

0%
1
0%
2
0%
3
0%
4

The normal to the curve,

${x}^{2}+2xy-{3y}^{2}=0,\ at\left (1,1\right)$ :

0%
Meets the curve, again in the fourth quadrant
0%
Does not meet the curve again
0%
Meets the curve again in the second quadrant
0%
Meets the curve again in the third quadrant

Number of tangents drawn from the point

$\left (-1/2,0\right)$ to the curve

$y={e}^{x}$ . (Here { } denotes fractional part function ).

0%
$2$
0%
$1$
0%
$3$
0%
$4$

Explanation

Let

$A$ be the point of contact of a tangent of

${e}^{x}$ passing through

$(-\cfrac{1}{2},0)$

$\Rightarrow$

${ \left| \cfrac { dy }{ dx } \right| }_{ \lambda }={ e }^{ \lambda }$

Equation of tangent

$\quad y-{ e }^{ \lambda }={ e }^{ \lambda }(x-\lambda )\quad$

$x=-\cfrac { 1 }{ 2 } ,y=0$

$-{ e }^{ \lambda }={ e }^{ \lambda }\left( -\cfrac { 1 }{ 2 } -\lambda \right) \Rightarrow { e }^{ \lambda }\lambda =\cfrac { { e }^{ \lambda } }{ 2 }$

either

$\quad { e }^{ \lambda }=0$ or

$\lambda =+\cfrac { 1 }{ 2 }$

$\quad y{ e }^{ \lambda }=0\Rightarrow y=0\quad \rightarrow$ xaxis

$\quad \lambda =\cfrac { 1 }{ 2 } \Rightarrow y-{ e }^{ \cfrac { 1 }{ 2 } }={ e }^{ \cfrac { 1 }{ 2 } }\left( x-\cfrac { 1 }{ 2 } \right)$

2 solutions or 2 distinct tangents

If the tangent at

$(x_{1}, y_{1})$ to the curve

$x^{3}+y^{3}=a^{3}$ meets the curve again at

$(x_{2}, y_{2})$ then

0%
$\dfrac{x_{2}}{x_{1}}+\dfrac{y_{2}}{y_{1}}=-1$
0%
$\dfrac{x_{2}}{y_{1}}+\dfrac{x_{1}}{y_{2}}=-1$
0%
$\dfrac{x_{1}}{x_{2}}+\dfrac{y_{1}}{y_{2}}=-1$
0%
$\dfrac{x_{2}}{x_{1}}+\dfrac{y_{2}}{y_{1}}=1$

Explanation

Given

$x^3+y^3=a^3$

The derivative is

$\dfrac{dy}{dx}=-\dfrac{x^2}{y^2}.....(1)$

Therefore, slope of tangent at

$(x_1,y_1)$ is

$-\dfrac{x_1^2}{y_1^2}.........(2)$

The tangent passes through

$(x_2,y_2)$ , therefore the slope of tangent is also given by

$\dfrac{y_2-y_1}{x_2-x_1}.......(3)$

Comparing the two slope equations we get

$\dfrac{y_2-y_1}{x_2-x_1}=-\dfrac{x_1^2}{y_1^2}$

$\dfrac{y_2^3-y_1^3}{x_2^3-x_1^3}\times\dfrac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}=-\dfrac{x_1^2}{y_1^2}$

$-\dfrac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}=-\dfrac{x_1^2}{y_1^2}$

$x_1^2y_1^2+x_1x_2y_1^2+x_2^2y_1^2=x_1^2y_1^2+x_1^2y_1y_2+x_1^2y_2^2$

$x_1x_2y_1^2+x_2^2y_1^2=x_1^2y_1y_2+x_1^2y_2^2$

$x_1^2y_2^2-x_2^2y_1^2=x_1x_2y_1^2-x_1^2y_1y_2$

$(x_1y_2-x_2y_1)(x_1y_2+x_2y_1)=x_1y_1(x_2y_1-x_1y_2)$

$x_1y_2+x_2y_1=-x_1y_1$

$\dfrac{x_2}{x_1}+\dfrac{y_2}{y_1}=-1$

If the tangent at P of the curve

$y^2=x^3$ intersects the curve again at Q and the straight lines OP, OQ ma angles

$\alpha, \beta$ with the x-axis where 'O' is the origin then

$\tan\alpha/\tan\beta$ has the value equal to?

0%
$-1$
0%
$-2$
0%
$2$
0%
$\sqrt{2}$

Explanation

Let

$P$ be

$(x_{1}, y_{1})$ and

$Q$ be

$(x_{2}, y_{2})$

equation of tangent

$(x_{2}, y_{2})$ , so

$\dfrac{y_{0}-y_{1}}{x_{2}-x_{1}}= \dfrac{3x_{1}^{2}}{2y_{1}}$

need to find out

$\dfrac{\tan \alpha}{ \tan \beta} = \dfrac{\dfrac{y_{1}}{x_{1}}}{y_{2}/x_{2}} = \dfrac{y_{1}x_{2}}{x_{1} y_{2}}$

$\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}= \dfrac{3x_{1}^{2}}{2y_{1}}$

$2y_{1}y_{2}-2y^{2}= 3x^{2}_{1}x_{2}-3x_{1}^{3}$

$2y_{1}y_{2}-2y_{1}^{2}= 3x_{1}^{2}x_{2}-3y^{2}$

$2y_{1}y_{2}+y_{1}^{2}= 3x_{1}^{2} x_{2}$

$y_{1}^{3} (2y_{2}+y_{1})^{3}= 27 x_{1}^{6} x^{3}_{2}$

$y_{1}^{3} (2y_{2}+ y_{1})^{3}= 27 y_{1}^{4} y_{2}^{3}$

$(2y_{2}+y_{1})^{3} - 27 y_{1} y_{2}$

$8 y_{2}^{3} - 15 y_{2}^{2} y_{1}+ 6y_{2} y^{2}_{1}+ y_{1}^{3}=0$

$(y_{2}-y_{1})^{2} (8y_{2}+ y_{1})=0$

$8y_{2}=-y_{1} \Rightarrow 64 y_{2}^{2} = y_{1}^{2}$

$64 x_{2}^{3} = x_{1}^{3} \Rightarrow 4 x_{2} = x_{1}$

$\dfrac{\tan \alpha}{\tan \beta}= \dfrac{y_{1}x_{2}}{x_{1}y_{2}}$

$= \dfrac{-8y_{2} x_{2}}{4x_{2} y_{2}} = -2$

A curve C has the property that if the tangent drawn at any point 'P' on C meets the coordinate axes at A and B, and P is midpoint of AB. If the curve passes through the point

$(1, 1)$ then the equation of the curve is?

0%
$xy=2$
0%
$xy=3$
0%
$xy=1$
0%
$xy=4$

Explanation

By option verification passing through

$(1,1)$

$xy=1$

The area of the triangle formed by the coordinate axes and a tangent to the curve

$xy={a}^{2}$ at the point

$({x}_{1},{y}_{1})$ is

0%
$\cfrac{{a}^{2}{x}_{1}}{{y}_{1}}$
0%
$\cfrac{{a}^{2}{y}_{1}}{{x}_{1}}$
0%
$2{a}^{2}$
0%
$4{a}^{2}$

Explanation

$\dfrac { dy }{ dx } =\dfrac { { -a }^{ 2 } }{ { x }^{ 2 } }$

${ x }_{ 1 }{ y }_{ 1 }={ a }^{ 2 }$

$y-{ y }_{ 1 }=\dfrac { { -a }^{ 2 } }{ { x }_{ 1 }^{ 2 } } \left( x-{ x }_{ 1 } \right)$

$x=0;y-{ y }_{ 1 }=\dfrac { { +a }^{ 2 } }{ { x }_{ 1 }^{ 2 } } \left( 0-{ x }_{ 1 } \right)$

$x=0;y={ y }_{ 1 }+\dfrac { { +a }^{ 2 } }{ { x }_{ 1 } }$

$y=0;{ +y }_{ 1 }=\dfrac { { +a }^{ 2 } }{ { x }_{ 1 }^{ 2 } } \left( x-{ x }_{ 1 } \right)$

$x={ x }_{ 1 }+\dfrac { { x }_{ 1 }^{ 2 }{ y }_{ 1 } }{ { a }^{ 2 } }$

Area

$=\dfrac { 1 }{ 2 } \left( { x }_{ 1 }+\dfrac { { x }_{ 1 }^{ 2 }{ y }_{ 1 } }{ { a }^{ 2 } } \right) \left( { y }_{ 1 }+\dfrac { { a }^{ 2 } }{ { x }_{ 1 } } \right)$

$=\dfrac { 1 }{ 2 } \left( { x }_{ 1 }{ y }_{ 1 }+{ a }^{ 2 }+\dfrac { { \left( { x }_{ 1 }{ y }_{ 1 } \right) }^{ 2 } }{ { a }^{ 2 } } +{ x }_{ 1 }{ y }_{ 1 } \right)$

$=\dfrac { 1 }{ 2 } \left( { a }^{ 2 }+{ a }^{ 2 }+\dfrac { { a }^{ 4 } }{ { a }^{ 2 } } +{ a }^{ 2 } \right)$

$=\dfrac { 4{ a }^{ 2 } }{ 2 } =2{ a }^{ 2 }$

$x={t}^{2}$ and

$y=2t$ , then equation of the normal at

$t=1$ is

0%
$x+y-3=0$
0%
$x+y-1=0$
0%
$x+y+1=0$
0%
$x+y+3=0$

Explanation

$x=t^2$ ,

$y=2t$

$y^2=4t^2=4x$

$t=1$

$y^2=4x$

$x=1,y=2$

$2y \dfrac{dy}{dx}=4 \rightarrow \dfrac{dy}{dx}=\dfrac{2y}{y}$

$y=2$

$\dfrac{dy}{dx}=1$

$-\dfrac{dx}{dy}=-1$

equation of normal

$y-2=-1(x-1)$

$y-2+x-1=0$

$x+y-3=0$

1080826_1172700_ans_88a1f0fce2274e10bbed45c08c5fdbce.png

The equation of the tangent to curve

$\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=2$ at the point (a, b) is

0%
$\frac{x}{a}-\frac{y}{b}=0$
0%
$\frac{x}{a}+\frac{y}{b}=2$
0%
$\frac{x}{a}-\frac{y}{b}=1$
0%
$\frac{x}{a}+\frac{y}{b}=0$

Explanation

We have,

$\sqrt{\dfrac{x}{a}}+\sqrt{\dfrac{y}{b}}=2\,\,.......\,\,\left( 1 \right)$

$\dfrac{\sqrt{x}}{\sqrt{a}}+\dfrac{\sqrt{y}}{\sqrt{b}}=2$

On differentiation and we get,

$\dfrac{1}{2\sqrt{x}\sqrt{a}}+\dfrac{1}{2\sqrt{y}\sqrt{b}}\dfrac{dy}{dx}=0$

$\dfrac{1}{2\sqrt{y}\sqrt{b}}\dfrac{dy}{dx}=-\dfrac{1}{2\sqrt{x}\sqrt{a}}$

$\dfrac{dy}{dx}=-\dfrac{\sqrt{y}\sqrt{b}}{\sqrt{x}\sqrt{a}}$

At the point $\left( a,\,b \right)$ and we get,

$\dfrac{dy}{dx}=-\dfrac{\sqrt{b}\sqrt{b}}{\sqrt{a}\sqrt{a}}$

$\dfrac{dy}{dx}=-\dfrac{b}{a}$

Equation of tangent is

$y-{{y}_{1}}=\dfrac{dy}{dx}\left( x-{{x}_{1}} \right)$

$y-b=-\dfrac{b}{a}\left( x-a \right)$

$ay-ab=-bx+ab$

$ay+bx=2ab$

$\dfrac{ay+bx}{ab}=2$

$\dfrac{y}{b}+\dfrac{x}{a}=2$

$\dfrac{x}{a}+\dfrac{y}{b}=2$

Hence, this is the answer.

The equation of tangent to the ellipse

$4{x^2} + 9{y^2} = 36$ at

$(3, - 2)\,is\,$

0%
$2x - 3y = 6$
0%
$3x - 2y = 13$
0%
$x + y = 1$
0%
$x - y = 5$

Explanation

The equation of tangent at

$(x_1,y_1)$ to the ellipse

$4{x^2} + 9{y^2} = 36$ is

$4xx_1+9yy_1=36$ ......(1).

Here

$x_1=3, y_1=-2$ .

Then the equaiton of the tangent will be

$4x.3+9y.(-2)=36$ or,

$2x-3y=6$ .

The equation of the normal to the curve

${x}^{4}=4y$ through the point

$(2,4)$ is

0%
$x+8y=34$
0%
$x-8y+30=0$
0%
$8x-2y=0$
0%
$8x+y=20$

Explanation

$\dfrac {m}{4}=y$

$\dfrac {dy}{dx}=\dfrac {4x^3}{4}=x^3$

$\dfrac {dy}{dx}|_{(2,4)} =2^3=8$

lquaton of normal

$\rightarrow y-y_1=\dfrac {-1}{dy/dx}(x-x_1)$

$y-4=\dfrac {-1}{8}(x-2)$

$8y-32=-x+2$

$x+8y=34$

The equation of tangents to the ellipse

${x^2} + 4{y^2} = 25$ at the point whose ordinate is 2, is

0%
$3x + 8y - 25 = 0$
0%
$-3x + 8y = - 25$
0%
$-3x - 8y = 25$
0%
$3x + 8y = 35$

Explanation

$Given$

${{x}^{2}}+4{{y}^{2}}=25$

$formula\,used$

$equation\text{ of tangent from point}\left( x_{1}^{{}},y_{1}^{{}} \right)to\,\,ellipse\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\,is$

$\frac{xx_{1}^{{}}}{{{a}^{2}}}+\frac{yy_{1}^{{}}}{{{b}^{2}}}=1$

${{x}^{2}}+4{{y}^{2}}=25$

$\Rightarrow \frac{{{x}^{2}}}{25}+\frac{4{{y}^{2}}}{25}=1$

$let\,the\,po\operatorname{int}\,be\,\left( x,2 \right)$

$hence$

$\frac{{{x}^{2}}}{25}+\frac{4{{\left( 1 \right)}^{2}}}{25}=1$

$\Rightarrow \frac{{{x}^{2}}}{25}=1-\frac{16}{25}$

$\Rightarrow {{x}^{2}}=9$

$\Rightarrow x=\pm 3$

$Hence\,the\,po\operatorname{int}\,is\,\left( -3,2 \right),\left( 3,2 \right)$

$equation\,of\,\tan gent\,at\left( 3,2 \right)$

$\frac{3x}{25}+\frac{4\left( 2y \right)}{25}=1$

$\Rightarrow 3x+8y=25$

The equation of tangent to the curve

$y=x^{2}+4x+1$ at

$\left(-1,-2\right)$ is

0%
$2x+y-5=0$
0%
$2x-y=0$
0%
$2x-y-1=0$
0%
$x+y-1=0$

Explanation

$\dfrac{dy}{dx}$ at

$(-1,-2)$ be:-

$\dfrac{d}{dx}(x^{2}+4x+1)=2x+4$

$2(-1)+4=2$

$\therefore$ Equation of tangent : -

$y -(-2)=2(x-(-1))$

$\therefore y +2=2x+2$

$\therefore y =2x$

$\therefore 2x-y=0$

Two lines drawn through the point

$A ( 4,0 )$ divide the area bounded by the curve

$y = \sqrt { 2 } \sin ( \pi x / 4 )$ and

$x$ - axis between the lines

$x = 2$ and

$x = 4$ into three equal parts. Sum of the slopes of the drawn lines is:

0%
$- 4 \sqrt { 2 } / \pi$
0%
$- \sqrt { 2 } / \pi$
0%
$- 2 \sqrt { 2 } / \pi$
0%
None

Explanation

Area bounded by

$y=\sqrt{2}\sin\left(\dfrac{\pi x}{4}\right)$ and

$x-axis$ between the lines

$x=2$ and

$x=4,$

$\Delta=\sqrt{2}\displaystyle\int_2^4\sin\dfrac{\pi x}{4}dx$

$=\left[-\dfrac{4\sqrt{2}}{\pi}.\cos\dfrac{\pi x}{4}\right]^4_2$

$=\dfrac{4\sqrt{2}}{\pi}\,unit^2$

Let the drawn lines are

$L_1:y-m_1(x-4)=0$ and

$L_2:y-m_2(x-4)=0,$ meeting the line

$x=2$ at the points

$A$ and

$B,$ respectively.

Clearly,

$A=(2,-2m_1);\,B=(2,-2m_2)$ [ Fig ]

Now,

$\Delta_{ACD}=\dfrac{\Delta}{3}\Rightarrow \dfrac{4\sqrt{2}}{3\pi}=\dfrac{1}{2}.2.(-2m_1)$

$\Rightarrow$

$m_1=-\dfrac{2\sqrt{2}}{3\pi}$

Also,

$\Delta_{BCD}=\dfrac{2\Delta}{3}$

$\Rightarrow$

$\dfrac{8\sqrt{2}}{3\pi}=\dfrac{1}{2}\times 2-2m_2$

$\Rightarrow$

$m_2=\dfrac{-4\sqrt{2}}{3\pi}$

$\Rightarrow$ Required sum

$=-\dfrac{2\sqrt{2}}{3\pi}+\left(\dfrac{-4\sqrt{2}}{3\pi}\right)$

$=\dfrac{-6\sqrt{2}}{3\pi}$

$=\dfrac{-2\sqrt{2}}{\pi}$

1481122_1180874_ans_e8e5f610c4d249289be00d9181cb35c5.png

The tangent to the curve

$2a^2y=x^3-3ax^2$ is parallel to the x-axis at the points

0%
$(0 , 0) , (2a, - 2a)$
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$(0 , 0) , (-2a, 2a)$
0%
$(0 , 0) , (-2a, - 2a)$
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$(2 , 2) , (0 , 0)$

Explanation

Let

$(x_1,y_1)$ represent the required points.

The slope of the

$x-$ axis is

$0$

Here

$2a^2y=x^3-3ax^2$ , since the points lies on the curve we get

$2a^2y_1=x_1^3-3ax_1^2 \quad ...(1)$

Consider

$2a^2y=x^3-3ax^2$

On differentiating both sides with respect to

$x$ , we get

$2a^2\dfrac{dy}{dx}=3x^2-6ax$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{3x^2-6ax}{2a^2}$

Slope of the tangent at

$(x_1,y_1)=\left(\dfrac{dy}{dx}\right)_{(x_1,y_1)}=\dfrac{3x_1^2-6ax_1}{2a^2}$

It is given that slope of the tangent at

$(x_1,y_1)=$ slope of the

$x-$ axis.

$\Rightarrow \dfrac{3x_1^2-6ax_1}{2a^2}=0$

$\Rightarrow 3x_1^2-6ax_1=0$

$\Rightarrow x_1(3x_1-6a)=0$

$\Rightarrow x_1=0$ or

$x_1=2a$

Also, from

$(1)$

$2a^2y_1=0$ or

$2a^2y_1=8a^3-12a^3$

$\Rightarrow y_1=0$ or

$y_1=-2a$

Thus, the required points are

$(0,0)$ and

$(2a,-2a)$

$x-2y+k=0$ is a common tangent to

$\displaystyle{ y }^{ 2 }=4x\quad \& \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { 3 } } =1\left( a>\sqrt { 3 } \right)$ , then the value of a, k and other common tangent are given by

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$a = 2$
0%
$a = -2$
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$x+2y+4 = 0$
0%
$k=4$

Explanation

Given:

$x-2y+k=0$

$\Rightarrow\,2y=x+k$

$\Rightarrow\,y=\dfrac{x}{2}+\dfrac{k}{2}$

Comparing with

$y=mx+c$ we get

$m=\dfrac{1}{2}$ and

$c=\dfrac{k}{2}$

$\therefore\,y=mx+c$ is tangent to

${y}^{2}=4ax$ where

$c=\dfrac{a}{m}$

$\Rightarrow\,{y}^{2}=4x,\,a=1$

$\Rightarrow\,c=\dfrac{1}{m}$

$\Rightarrow\,\dfrac{k}{2}=\dfrac{1}{\dfrac{1}{2}}$

$\Rightarrow\,\dfrac{k}{2}=2$

$\therefore\,k=4$

Substituting

$m=\dfrac{1}{2}$ and

$c=\dfrac{1}{m}=\dfrac{1}{\dfrac{1}{2}}=2$ in

$y=mx+c$ we get

$\Rightarrow\,y=\dfrac{x}{2}+\dfrac{4}{2}$

$\Rightarrow\,y=\dfrac{x}{2}+2$

Given:

$y=mx+c$ is tangent to

$\dfrac{{x}^{2}}{{a}^{2}}+\dfrac{{y}^{2}}{{b}^{2}}=1$

$\therefore\,$ Required condition is

${c}^{2}={a}^{2}{m}^{2}+{b}^{2}$

$\Rightarrow\,4={a}^{2}{\left(\dfrac{1}{2}\right)}^{2}+3$

$\Rightarrow\,\dfrac{{a}^{2}}{4}=4-3=1$

$\Rightarrow\,{a}^{2}=4$

$\Rightarrow\,a=2$ since

$a>0$

The point on the curve

${x}^{2}+{y}^{2}-2x-3=0$ at which the tangent in parallel to x-axis is

0%
$(1,0),(-1,-4)$
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$(0,-1),(-2,3)$
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$(2,13),(-2,-3)$
0%
$(1,2),( 1,-2)$

Explanation

$(1,2)$ and

$(1,-2)$ are two points where the tangent is parallel to the x-axis.

The equation of curve can be reduced to

$(x-1)^2+y^2=4$

Therefore the center is

$(1.0)$ on the x-axis and the radius is

$2$ units

Now if we add and subtract

$2$ to the

$y$ co-ordinate of the center we get

$(1,2)$ and

$(1,-2)$ are two points where tangent is possible to x-axis.

$y=2$ and

$y=-2$ are the two equations of the tangent to the circle and also parallel to axis and equation of the line

$x=1$ represent diameter perpendicular to

$x-axis$

Slope of tangent to the circle

$( x - r ) ^ { 2 } + y ^ { 2 } = r ^ { 2 }$ at the point

$( x , y )$ lying on the circle is

0%
$\frac { x } { y - r }$
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$\frac { r - x } { y }$
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$\frac { y ^ { 2 } - x ^ { 2 } } { 2 x y }$
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$\frac { y ^ { 2 } + x ^ { 2 } } { 2 x y }$

Explanation

$(x-r)^2+y^2=r^2$

$\dfrac{d}{dx}(x-r)^2+\dfrac{d}{dx}(y^2)=\dfrac{d}{dx}(r^2)$

$2(x-r)+2y\dfrac{dy}{dx}=0$

$x-r+y\dfrac{d}{dx}=0$

Therefore the slope of the tangent is,

$\dfrac{dy}{dx}=\dfrac{r-x}{y}$

If the slope of one of the lines represented

${a^3}{x^2} + 2hxy + {b^3}{y^2} = 0$ be the square of the other, then

$ab(a+b)$ is equal to:

0%
$2h$
0%
$-2h$
0%
$8h$
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$-8h$

The normal to the curve,

${{\text{x}}^{\text{2}}}{\text{ + 2xy - 3}}{{\text{y}}^{\text{2}}}{\text{ = 0,}}\;{\text{at}}\;\left( {{\text{1,1}}} \right){\text{:}}$

0%
meets the curve again in the second quadrant.
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meets the curve again in the third quadrant.
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meets the curve again in the fourth quadrant.
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does not meet the curve again.

Explanation

Given equation of curve is

${x}^{2}+2xy-3{y}^{2}=0$ .......

$(1)$

On differentiating w.r.t

$x$ we get

$2x+2x{y}^{\prime}+2y-6y{y}^{\prime}=0$

$\Rightarrow\,{y}^{\prime}=\dfrac{x+y}{3y-x}$

$x=1,\,y=1,\,{y}^{\prime}=1$

$\left(\dfrac{dy}{dx}\right)_\left(1,1\right)=1$

Equation of the normal at

$\left(1,1\right)$ is

$y-1=-\dfrac{1}{1}\left(x-1\right)$

$\Rightarrow\,y-1=-x+1$

$\Rightarrow\,x+y=2$

$\Rightarrow\,y=2-x$ ......

$(2)$

On solving equations

$(1)$ and

$(2)$ ,we get

${x}^{2}+2x\left(2-x\right)-3{\left(2-x\right)}^{2}=0$

$\Rightarrow\,{x}^{2}+4x-2{x}^{2}-3\left(4+{x}^{2}-4x\right)=0$

$\Rightarrow\,{x}^{2}+4x-2{x}^{2}-12-3{x}^{2}+12x=0$

$\Rightarrow\,-4{x}^{2}+16x-12=0$

$\Rightarrow\,{x}^{2}-4x+3=0$

$\Rightarrow\,\left(x-1\right)\left(x-3\right)=0$

$\Rightarrow\,x=1,\,x=3$

When

$x=1,$ then

$y=1$ using

$(2)$

When

$x=3$ then

$y=-1$

$\therefore\,P\left(1,1\right)$ and

$Q\left(3,-1\right)$

Hence, normal meets the curve again at

$\left(3, - 1\right)$ in fourth quadrant.

The line

$3x-4y=0$

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is a tangent to the circle ${x}^{2}+{y}^{2}=25$
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is a normal to the circle ${x}^{2}+{y}^{2}=25$
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does not meet the circle ${x}^{2}+{y}^{2}=25$
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does not pass thro' the origin

Explanation

$3x=4y$

$x=\dfrac { 4y }{ 3 }$

${ x }^{ 2 }+{ y }^{ 2 }=25$

${ x }^{ 2 }-\left( 1+\dfrac { 16 }{ 9 } \right) =25$

${ x }^{ 2 }=9$

$x=\pm 3$

$y=\pm 9/4$

Now,

${ x }^{ 2 }+{ y }^{ 2 }=25$

$2x+2y\dfrac { dy }{ dx } =0$

$\dfrac { dy }{ dx } =\dfrac { -x }{ y } =\dfrac { -\left( 3 \right) }{ \left( 9/4 \right) } =-4/3$

$\dfrac { -dx }{ dy } =3/4\Rightarrow$ slope of normal at circle and

$3/4\Rightarrow$ solpe of

$3x-4y=0$

So, Line is normal at circle.

The curves

$x = y^2 and. xy = k$ cut at right angles, If

$6k^2$ = 1.

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True
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False

The equation of the normal to the curve

$y^4=ax^3$ at (a , a) is

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x + 2y=3a
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3x-4y+a=0
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4x+3y=7a
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4x-3y=a

Explanation

The equation of the curve is,

${ y }^{ 4 }=a{ x }^{ 3 }$

Differentiate w.r.t x, we get,

$4{ y }^{ 3 }\dfrac { dy }{ dx } =3a{ x }^{ 2 }$

$\therefore \dfrac { dy }{ dx } =\dfrac { 3a{ x }^{ 2 } }{ 4{ y }^{ 3 } }$

$\therefore { \dfrac { dy }{ dx } }_{ \left( a,a \right) }=\dfrac { 3a\left( a \right) ^{ 2 } }{ 4\left( a \right) ^{ 3 } }$

$\therefore { \dfrac { dy }{ dx } }_{ \left( a,a \right) }=\dfrac { 3\left( a \right) ^{ 3 } }{ 4\left( a \right) ^{ 3 } }$

$\therefore { \dfrac { dy }{ dx } }_{ \left( a,a \right) }=\dfrac { 3 }{ 4 }$

Thus, slope of tangent is

$\dfrac { 3 }{ 4 }$

Let

${ m }_{ 1 }$ = slope of normal

As tangent and normal are perpendicular to each other, we can write,

$\dfrac { 3 }{ 4 } \times { m }_{ 1 }=-1$

$\therefore { m }_{ 1 }=\dfrac { -4 }{ 3 }$

Thus, equation of normal passing through

$\left( a,a \right)$ is,

$y-{ y }_{ 1 }={ m }_{ 1 }\left( x-{ x }_{ 1 } \right)$

$\therefore y-a=\dfrac { -4 }{ 3 } \left( x-a \right)$

$\therefore 3\left( y-a \right) =-4\left( x-a \right)$

$\therefore 3y-3a=-4x+4a$

$\therefore 4x+3y=7a$

Which of the following lines, is a normal to the parabola

${y}^{2}=16x$ ?

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$y=x-11\cos{\theta}-3\cos{3\theta}$
0%
$y=x-11\cos{\theta}-\cos{3\theta}$
0%
$y=(x-11)\cos{\theta}+\cos{3\theta}$
0%
$y=(x-11)\cos{\theta}-\cos{3\theta}$

Explanation

$\begin{array}{l} Accoding\, to\, the\, \, question, \\ Equ\, of\, parabola\, is........ \\ y=mx-a{ m^{ 3 } }-2am\, \, \, \, \, \, \, (where\, \, m=slope,\, \, condition\, of\, normal\, \, is,\, c=-2am-a{ m^{ 3 } }) \\ given, \\ y=(x-11)\cos \theta -\cos 3\theta \\ \, \, \, \, =x\cos \theta -11\cos \theta -\cos 3\theta \, \, \, \, \, (where\, m=\cos \theta ) \\ \, \, \, \, c=-11\cos \theta -\cos 3\theta \\ \, \, \, =-11\cos \theta -(4co{ s^{ 3 } }\theta -3\cos \theta ) \\ \, \, \, =\, -8\cos \theta -4{ \cos ^{ 3 } }\theta \, is\, the\, required\, option \\ So,the\, correct\, option\, is\, D. \end{array}$

The equation of one of the tangents to the curve

$y=\cos(x+y),-2\pi \le x \le 2\pi$ ; that is parallel to the line

$x+ 2y = 0$ , is

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$x+2y=1$
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$x+2y=\dfrac {\pi}{2}$
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$x+2y=\dfrac {\pi}{4}$
0%
$None\ of\ these$

Explanation

Given curve is

$y = \cos (x + y), -2\pi \leq x \leq 2\pi ..... (1)$

Parallel to the line

$x + 2y = 0$

We prove that, slope of tangent is

$\dfrac {dy}{dx}$

$y = \cos (x + y)$

Diff w.r.t

$x$ and we get

$\dfrac {dy}{dx} = -\sin (x + y) \dfrac {d}{dx}(x + y) \therefore \dfrac {d}{dx}\cos x = -\sin x$

$\dfrac {dy}{dx} = -\sin (x + y) \left (1 + \dfrac {dy}{dx}\right )$

$\dfrac {dy}{dx} = -\sin (x + y) + (-\sin (x + y))\dfrac {dy}{dx}$

$\dfrac {dy}{dx} [1 + \sin (x + y)] = -\sin (x + y)$

$\dfrac {dy}{x} = \dfrac {-\sin (x + y)}{1 + \sin (x + y)} ..... (2)$

Given line is

$x + 2y = 0$

$x = -2y$

$2y = -x$

$y = -\dfrac {x}{2}$

$y = \dfrac {-x}{2} + C$

The above equation of the form

$y = mx + C$

When,

$m = \dfrac {-1}{2}$ (slope)$$

$\dfrac {dy}{dx} = m = \dfrac {-1}{2}$

$\therefore$ slope of tangent

$=$ slope of line

$\dfrac {-\sin (x + y)}{1 + \sin (x + y)} = -\dfrac {1}{2}$

$\dfrac {\sin (x + y)}{1 + \sin (x + y)} = \dfrac {1}{2}$

$2\sin (x + y) = 1 + \sin (x + y)$

$2 \sin (x + y) - \sin (x + y) = 1$

$\sin (x + y) = 1$

$\sin (x + y) = \sin \dfrac {\pi}{2}$

Hence, general solution of

$x + y$ is

$x + y = n\pi + (-1)^{n} \dfrac {\pi}{2}$

Now, Finding points through which tangents passes.

$y = \cos (x + y)\rightarrow$ given curve

Putting value of

$(x + y)$

$y = \cos \left [n\pi + (1)^{n}\dfrac {\pi}{2}\right ]$

$y = 0$ in

$x$ (put)

$x + y = n\pi + (1)^{n}\dfrac {\pi}{2}$

$x + 0 = n\pi + (-1)^{n} \dfrac {\pi}{2}$

$x = n\pi + (-1)^{n} \dfrac {\pi}{2}$

Since,

$-2\pi \leq x \leq 2\pi$

Putting

$n = 0$

$x = 0\pi + (-1)^{0} \dfrac {\pi}{2}$

$\therefore (-1)^{0} = 1$

$x = \dfrac {\pi}{2}$

Putting

$n = -1$

$x= (-1)\pi + (-1)^{-1} \dfrac {\pi}{2}$

$x = -\pi - \dfrac {\pi}{2}$

$x = -\dfrac {3\pi}{2}$

Thus, points are

$\left (-\dfrac {3\pi}{2}, 0\right )$ and

$\left (\dfrac {\pi}{2}, 0\right )$

We know that equation of tangent at point

$(x_{1}, y_{1})$ and slope

$m$ is

$y - y_{1} = m(x - x_{1})$

Now, Equation of tangent at

$\left (\dfrac {-3\pi}{2}, 0\right )$ and slope

$\dfrac {-1}{2}$ is

$y - 0 = -\dfrac {1}{2} \left (x - \left (\dfrac {-3\pi}{2}\right )\right )$

$y = \dfrac {-1}{2} \left (x + \dfrac {3\pi}{2}\right )$

$4y = -(2x + 3\pi)$

$2x + 4y + 3\pi = 0$

Equation of tangent at

$\left (\dfrac {\pi}{2}, 0\right )$ and slope

$\dfrac {-1}{2}$ is

$y - 0 = \dfrac {-1}{2} \left (x - \dfrac {\pi}{2}\right )$

$y = \dfrac {-1}{2} \left (\dfrac {2x - \pi}{2}\right )$

$4y = -2x + \pi$

$2x + 4y - \pi = 0$

Hence, this is the answer.

The tangent at any point of the curve

$x={ at }^{ 3 },y={ at }^{ 4 }$ divides the abscissa of the point of contact in the ratio

0%
$1:4$
0%
$3:2$
0%
$1:3$
0%
$3:1$

Explanation

Given:

$x = at^2$

$y = at^4$

Then,

$\Rightarrow t^3 = \dfrac{x}{a}, t^4= \dfrac{y}{a}$

$\Rightarrow t = \left(\dfrac{x}{a}\right)^{1/3}\, t = \left(\dfrac{y}{a}\right)^{1/4}$

$\Rightarrow \left(\dfrac{x}{a}\right)^{1/3} = \left(\dfrac{y}{a}\right)^{1/4}$

$\Rightarrow \left(\dfrac{x}{a}\right)^{\frac{1}{3}\times \frac{4}{12}} = \left(\dfrac{y}{a} \right)^{\frac{1}{4}\times 12}$

$\Rightarrow \dfrac{x^4}{a^4} = \dfrac{y^3}{a^2}$

$\Rightarrow y^3 = \dfrac{x^4}{a}$ at

$P(h,k)$

Also relation

$k^3 = \dfrac{h^4}{a}$

Now by differentiation of after equation we get

$3y^2 \dfrac{dy}{dx} = \dfrac{4x^2}{a}$

$\dfrac{dy}{dx} = \dfrac{4x^3}{3ay^2}$

Now

$M_T = \dfrac{dy}{dx} = \dfrac{4x^3}{3ay^2}$

Now we will find slope of tangent

$M_T / P(h,x) = \dfrac{4h^3}{3ax^2}$

equation of tangent

$P(h,k)$

$y - k = \dfrac{4h^3}{3ak^2}(x-h)$

$\Rightarrow$ Let

$y = 0$

Then,

$-k = \dfrac{4h^3}{3ak^2}(x-h)$

$\Rightarrow \dfrac{-3ak^3}{4h^3} = x - h$

Earlier we found that

$k^3 = \dfrac{h^4}{a}$

Then

$ax^3 = h^4$

$\dfrac{-3h^4}{4h^2} = x-h$

$\Rightarrow \dfrac{-3h}{4}+h = x$

$\Rightarrow x = \dfrac{h}{4}$

$\Rightarrow \dfrac{x}{h} = \dfrac{1}{4}$

Hence option (a)

$1 : 4$ is the correct choice

1496738_1244227_ans_0a282e0fbce94a80bd120e501512f784.JPG

State true or false.

The curves

$y = {x^2} - 3x + 1$ and

$x\left( {y + 3} \right) = 4$ intersect at the right angles at their point of intersection

0%
True
0%
False

Explanation

First curve is given by,

$y={ x }^{ 2 }-3x+1$ (1)

Differentiate w.r.t. x, we get,

$\frac { dy }{ dx } =2{ x }-3$

Thus, slope of tangent to this curve is,

${ m }_{ 1 }=2{ x }-3$

Second curve is given by,

$x\left( y+3 \right) =4$

$\therefore \left( y+3 \right) =\frac { 4 }{ x }$ (2)

Differentiate w.r.t. x, we get,

$\frac { dy }{ dx } =-\frac { 4 }{ { x }^{ 2 } }$

Thus, slope of tangent to this curve is,

${ m }_{ 2 }=-\frac { 4 }{ { x }^{ 2 } }$

1) Point of intersection of two curves-

Equation of first curve is given by,

$y={ x }^{ 2 }-3x+1$

Equation of second curve is given by,

$y=\frac { 4 }{ x } -3$

Thus, we can equate these equations, we get,

${ x }^{ 2 }-3x+1=\frac { 4 }{ x } -3$

Multiply both sides by x, we get,

$\therefore { x }^{ 3 }-3{ x }^{ 2 }+x=4-3x$

$\therefore { x }^{ 3 }-3{ x }^{ 2 }+4x-4=0$

Put

$x=2$ in above equation,

$LHS={ \left( 2 \right) }^{ 3 }-3{ \left( 2 \right) }^{ 2 }+4\left( 2 \right) -4$

$=8-\left( 3\times 4 \right) +8-4$

$=8-12+8-4$

$=0=RHS$

Thus,

$x=2$ is one factor of given equation.

By synthetic division, we can get quadratic equation as,

${ x }^{ 2 }-x+2=0$

$\therefore x=\dfrac { -\left( -1 \right) \pm \sqrt { { \left( -1 \right) }^{ 2 }-4\times 1\times 2 } }{ 2\times 1 }$

$\therefore x=\dfrac { 1\pm \sqrt { 1-8 } }{ 2 }$

$\therefore x=\dfrac { 1\pm \sqrt { -7 } }{ 2 }$

These roots will be imaginary and are neglected.

$\therefore x=2$

Put this value in equation (1), we get,

$y={ \left( 2 \right) }^{ 2 }-3\left( 2 \right) +1$

$\therefore y=4-6+1$

$\therefore y=-1$

Thus, point of intersection of two curves is,

$\left( 2,-1 \right)$

${ \left[ { m }_{ 1 } \right] }_{ \left( 2,-1 \right) }=2\left( 2 \right) -3$

$\therefore { \left[ { m }_{ 1 } \right] }_{ \left( 2,-1 \right) }=1$

${ \left[ { m }_{ 2 } \right] }_{ \left( 2,-1 \right) }=\dfrac { -4 }{ { \left( 2 \right) }^{ 2 } }$

$\therefore { \left[ { m }_{ 2 } \right] }_{ \left( 2,-1 \right) }=\dfrac { -4 }{ 4 } =-1$

Now,

${ m }_{ 1 }\times { m }_{ 2 }=1\times -1$

$\therefore { m }_{ 1 }\times { m }_{ 2 }=-1$

Thus, curves intersect each other at right angle

The slope of the straight line which is both tangent and normal to the curve

$4x^3=27y^2$ is

0%
$\underline { + } 1$
0%
$\underline { + } \dfrac { 1 }{ 2 }$
0%
$\underline { + } \dfrac { 1 }{ \sqrt { 2 } }$
0%
$\underline { + } \sqrt { 2 }$

Explanation

$Given:$

$4{{x}^{3}}=27{{y}^{2}}$

$differentiating\,w.r.t.x$

$12{{x}^{2}}=54yy'$

$t$

$The\text{ line is tangent at}\,\text{P}\left( 3{{t}^{2}},2{{t}^{3}} \right)and\,normal\,at\left( 3t_{1}^{2},2t_{1}^{3} \right)$

$\frac{dy}{dx}=\frac{2\left( 9{{t}^{4}} \right)}{9\left( 2{{t}^{3}} \right)}=1$

$equation\,to\,\tan gent\,P\left( t \right)is$

$y-2{{t}^{3}}=t\left( x-3{{t}^{2}} \right)$

$\Rightarrow tx-y={{t}^{3}}\,\,\,\,\,........\left( i \right)$

$Equation\,to\,normal\,atQ\left( t_{1}^{{}} \right)is$

$y-2t_{1}^{3}=-\frac{1}{t_{1}^{{}}}\left( x-3t_{1}^{2} \right)$

$\Rightarrow x+t_{1}^{{}}y=2t_{1}^{4}+3t_{1}^{2}\,\,\,\,\,\,\,\,...........\left( ii \right)$

$\left( i \right)\,and\left( ii \right)are\,identical$

$\frac{t}{1}=\frac{-1}{t_{1}^{{}}}=\frac{{{t}^{3}}}{2t_{1}^{4}+3t_{1}^{2}}$

$\Rightarrow {{t}^{2}}=\frac{2}{{{t}^{4}}}+\frac{3}{{{t}^{2}}}\,\,\,\,\,\,\,\,or\,\,t_{1}^{{}}=-\frac{1}{t}$

$\Rightarrow {{t}^{6}}=2+3{{t}^{2}}$

$\Rightarrow {{t}^{6}}-3{{t}^{2}}-2=0$

$\Rightarrow {{t}^{2}}=2$

$t=\pm \sqrt{2}$

The angle between the curves

$y = \sin x$ and

$y = \cos x$ is

0%
$\tan^{-1}(2\sqrt{2})$
0%
$\tan^{-1}(3\sqrt{2})$
0%
$\tan^{-1}(3\sqrt{3})$
0%
$\tan^{-1}(5\sqrt{2})$

Explanation

C₁ :

$y = \sin x$ , C₂ :

$y = \cos x$ .
Equating the

$y$ ' s,

$\sin x = \cos x$ ∴

$x = \dfrac{\pi}{4}$
∴ curves intersect each other at the point P :

$x = \dfrac{\pi}{4}$ .
Now, differentiating w.r.t.

$x$ ,

C₁ gives :

$\dfrac{dy}{dx} = \cos x$

C₂ gives :

$\dfrac{dy}{dx} = - \sin x$ .

Hence slopes

$m₁ and m₂ of C₁ and C₂ at P :$

$x = \dfrac{π}{4}$

$are m₁$

$= \cos \dfrac{π}{4} = \dfrac{1}{√2}$ $

m₂ =

$- \sin \dfrac{π}{4} = -\dfrac{1}{√2}$ .

$θ$ is the acute angle between them at

$P$ , then

$tan θ =\dfrac{ | ( m₁ - m₂ )|}{ |( 1 + m₁m₂ ) |}$

$=\dfrac{ | ((\dfrac{1}{√2}) - (-\dfrac{1}{√2}))| }{ |( 1 + (\dfrac{1}{√2})(-\dfrac{1}{√2})) | }$

$=\dfrac{ | 2( \dfrac{1}{√2 })|}{ |( \dfrac{(2-1)} { (2)} | }$

$= 2√2$
∴

$θ = \tanֿ¹ ( 2√2 )$

The normal to the curve

$x=\quad a(cos\theta +\theta sin\theta ),\quad y=\quad a(sin\theta -\theta cos\theta )$ at any point

$'\theta '$ is such that

0%
it passes through the origin
0%
it makes an angle $\dfrac { \pi }{ 2 } +\theta$ with the x-axis
0%
it is at a constant distance from the origin
0%
it passes through $\left(a,\dfrac{\pi}{2}\right)$

Explanation

$\dfrac{dy}{dx}=\dfrac{dy}{d\theta}.\dfrac{d\theta}{dx}=\tan{\theta}=$ Slope of tangent

$\therefore$ Slope of normal to the curve

$=-\cot{\theta}=\tan{\left(90+\theta\right)}$

Now, equation of normal to the curve

$\left[y-a\left(\sin{\theta}-\theta\cos{\theta}\right)\right]=\dfrac{-\cos{\theta}}{\sin{\theta}}\left(x-a\left(\cos{\theta}+a\sin{\theta}\right)\right)$

$\Rightarrow x\cos{\theta}+y\sin{\theta}=a(1)$

Now, distance from

$\left(0,0\right)$ to

$x\cos{\theta}+y\sin{\theta}=a$ is

distance

$=d=\dfrac{0+0-a}{1}=a$

$\therefore$ Distance is constant

$=a$

The tangent to the curve

$y=e^{2x}$ at the point

$(0, 1)$ meets x-axis at?

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$(0, 1)$
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$\left(-\dfrac{1}{2}, 0\right)$
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$(2, 0)$
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$(0, 2)$

Explanation

solution - equation of curve is y=

$e^{2x}$

tangent to the curve is

$\frac{dy}{dx}=2e^{2x}$

$(\frac{de^{x}}{dx}=e^{x})$

At points (0,1), the tangent is

$(\frac{dy}{dx})_{(0,1)}=2e^{2(0)}$

$(\frac{dy}{dx})_{(0,1)}=2e^{0}$

$[e^{0}=1]$

$\frac{dy}{dx}_{0,1}=2$

when the slope is 2, and solve the tangent

meets y axis the coordinate is 0,

Hence the tangent meets the x axis at (2,0)

1176120_1294135_ans_11a7c6cc3b38459bb3209773aa2a4a3f.jpg

Three normals are drawn from the point

$\left(c,0\right)$ to the curve

${y}^{2}=x.$ If two of the normals are perpendicular to each other,then

$c=$

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$\dfrac{1}{4}$
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$\dfrac{1}{2}$
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$\dfrac{3}{4}$
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$1$

Explanation

$Let\,the\text{ equation of normal to }{{\text{y}}^{2}}=4ax\,is$

$y=mx-2am-a{{m}^{2}}$

$\therefore equation\,of\,normal\,for\,{{y}^{2}}=x\,is$

$y=mx-\frac{m}{2}-\frac{1}{4}{{m}^{3}}which\,passes\,through\left( c,0 \right)$

$\therefore 0=m\left( c-\frac{1}{2}-\frac{{{m}^{2}}}{4} \right)$

$\Rightarrow m=0\,\,and\,\frac{{{m}^{2}}}{4}=c-\frac{1}{2}$

$\Rightarrow m=\pm 2\sqrt{c-\frac{1}{2}}$

$\text{which gives a normal as x-axis and for other two normal}$

$c-\frac{1}{2}>0$

$\Rightarrow c>\frac{1}{2}$

$N\text{ow, if normals are perpendicular}\text{.}$

$\left( 2\sqrt{c-\frac{1}{2}} \right)\left( -2\sqrt{c-\frac{1}{2}} \right)=-1$

$\Rightarrow c-\frac{1}{2}=\frac{1}{4}$

$\Rightarrow c=\frac{3}{4}$

The equation to the normal to the curve

$y=\sin x$ at

$(0, 0)$ is

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$x=0$
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$y=0$
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$x+y=0$
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$x-y=0$

Explanation

Given:

$y=\sin x$

To find: Eq. of normal to curve at

$(0, 0)$

Solution:

$y=\sin x$

$\dfrac{dy}{dx}=\cos x$

$\left.\dfrac{dy}{dx}\right]_{(0, 0)}=\cos 0=1$

Eq. of normal:

Slope of normal

$=\dfrac{-1}{\dfrac{dy}{dx}}=-1$

Eq. of normal at

$(0, 0)$

$\Rightarrow y-0=-1(x-0)$

$y=-x$

$y+x=0$

$\therefore x+y=0$ is the required.

Eq. of normal to curve

$y=\sin x$ at

$(0, 0)$ .

The tangent to the curve,

$y = xe^{x^2}$ passing through the point

$(1, e)$ also passes through the point:

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$\left(\dfrac{4}{3}, 2e\right)$
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$(2, 3e)$
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$\left(\dfrac{5}{3}, 2e\right)$
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$(3, 6e)$

Explanation

$\textbf{Step 1: Find slope of the tangent to the given curve at point (1,e).}$

$\text{Given curve : }\ y=xe^{x^{2}}.........(1)$

$\text{Differentiating equation (1) with respect to }\ x,$

$\Rightarrow \dfrac{dy}{dx} = x\cdot e^{x^2} \cdot 2x+e^{x^2}$

$\text{Slope of tangent at point } {(1,e)} \text{ is }$

$m=\dfrac{dy}{dx}\left|_{(1,e)} = \left(x\cdot e^{x^2} \cdot 2x+e^{x^2}\right)\right|_{(1,e)} = 2\cdot e + e = 3e$

$\textbf{Step 2: Find the equation of the tangent}$

$\textbf{Using one point slope form of equation of the line: } \boldsymbol{ y-y_{1}=m(x-x_{1})}$

$\text{Equation of tangent is: }$

$y - e = 3e (x - 1)$

$\boldsymbol{[\because m=2e]}$

$\Rightarrow y = 3ex - 3e + e$

$\Rightarrow y = (3e) x - 2e............(2)$

$\textbf{Step 3: To find correct answer, check by given options}$

$\left(\dfrac{4}{3}, 2e\right)$

$\text{lies on it, as putting this point in (2) we get,}$

$2e=3e\times \dfrac 43-2e\Rightarrow 2e=2e$

$\Rightarrow \text{L.H.S = R.H.S}$

$\textbf{Hence, the correct answer is option 'A'.}$

If the line

$x+y=0$ touches the curve

$2y^2=\alpha x^2+\beta$ at

$(1,-1),$ then

$(\alpha ,\beta )=$

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$(-2,4)$
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$(-1,3)$
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$(4,-2)$
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$(2,0)$

Explanation

Given equation of curve is,

$2{ y }^{ 2 }=\alpha { x }^{ 2 }+\beta$

Point

$\left( 1,-1 \right)$ lies on the curve. Thus, it must satisfy given equation of curve.

$\therefore 2{ \left( -1 \right) }^{ 2 }=\alpha { \left( 1 \right) }^{ 2 }+\beta$

$\therefore 2\left( 1 \right) =\alpha { \left( 1 \right) }+\beta$

$\therefore 2=\alpha +\beta$ (1)

Now, equation of tangent to curve is,

$x+y=0$

$\therefore y=-x$

Thus, slope of tangent is,

${ m }_{ 1 }=-1$ (2)

Equation of the curve is,

$2{ y }^{ 2 }=\alpha { x }^{ 2 }+\beta$

Differentiate w.r.t. x, we get,

$2\times 2y\frac { dy }{ dx } =2\alpha x+0$

$\therefore 4y\frac { dy }{ dx } =2\alpha x$

$\therefore \frac { dy }{ dx } =\frac { 2\alpha x }{ 4y }$

Slope of curve at

$\left( 1,-1 \right)$ is,

${ m }_{ 2 }=\frac { 2\alpha \times 1 }{ 4\times -1 }$

${ m }_{ 2 }=\frac { 2\alpha }{ -4 }$

$\therefore { m }_{ 2 }=\frac { \alpha }{ -2 }$ (3)

At point

$\left( 1,-1 \right)$ , slope of tangent and normal will be same.

Thus, from equation (2) and (3),

$\frac { \alpha }{ -2 } =-1$

$\therefore \alpha =2$

Put this value in equation (1), we get,

$2+\beta =2$

$\therefore \beta =0$

$\therefore \left( \alpha ,\beta \right) =\left( 2,0 \right)$

Equation of the tangent at (1, -1) to the curve

${ x }^{ 3 }-x{ y }^{ 2 }-4{ x }^{ 2 }-xy+5x+3y+1=0$ is

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$x-4y-5=0$
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$x+1=0$
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$y-1=0$
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$y+1=0$

The angle made by the tangent at any point on the curve

$x=a(t+\sin { t } \cos { t } ),y=a{ (1+\sin { t } ) }^{ 2 }$ with x-axis is

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$\dfrac { \pi }{ 2 }$
0%
$\dfrac { \pi }{ 4 }$
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$\pi +\dfrac { t }{ 2 }$
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$\dfrac { \pi }{ 4 } +\dfrac { t }{ 2 }$

Explanation

Given,

$x=a(t+\sin t \cos t)$

$\dfrac{dx}{dt}=\dfrac{d}{dt}a(t+\sin t \cos t)$

$\therefore \dfrac{dx}{dt}=a[1+\cos ^2t-\sin ^2t]=a\left(1+\cos \left(2t\right)\right)=2a\cos ^2t$

$y=a(1+\sin t)^2$

$\dfrac{dy}{dt}=\dfrac{d}{dt}a(1+\sin t)^2$

$\therefore \dfrac{dy}{dt}=2a\left(1+\sin \left(t\right)\right)\cos \left(t\right)$

$\Rightarrow \dfrac{dy}{dx}$

$=\dfrac{2a\left(1+\sin \left(t\right)\right)\cos \left(t\right)}{2a\cos ^2t}$

$=\dfrac {1+\sin t}{\cos t}$

$=\dfrac{\cos \frac{t}{2}+\sin \frac{t}{2}}{\cos \frac{t}{2}-\sin \frac{t}{2}}$

$=\dfrac{1+\tan \frac{t}{2}}{1 -\tan \frac{t}{2}}$

$=\tan \left ( \dfrac{\pi }{4}+\dfrac{t}{2} \right )$

angle made by tangent,

$\tan \theta =\tan \left ( \dfrac{\pi }{4}+\dfrac{t}{2} \right )$

$\Rightarrow \theta =\dfrac{\pi }{4}+\dfrac{t}{2}$

Length of the normal to the curve at any point on the curve

$y=\dfrac { a\left( { e }^{ x/a }+{ e }^{ -x/a } \right) }{ 2 }$ varies as

0%
$x$
0%
${ x }^{ 2 }$
0%
$y$
0%
${ y }^{ 2 }$

Explanation

Given,

$y=\dfrac{a(e^{\frac{x}{a}}+e^{-\frac{x}{a}})}{2}$

$\Rightarrow y=a\cos h\left ( \dfrac{x}{a} \right )$

$\dfrac{dy}{dx}=a \sin h\left ( \dfrac{x}{a} \right )\dfrac{1}{a}$

$=\sin h\left ( \dfrac{x}{a} \right )$

Length of normal

$=y\left [ 1+\left (\dfrac{dy}{dx} \right )^2 \right ]^{\frac{1}{2}}$

$=a\cos h\left ( \dfrac{x}{a} \right )\left [ 1+ \sin ^2h\left ( \dfrac{x}{a} \right )\right ]^{\frac{1}{2}}$

$=a\cos h\left ( \dfrac{x}{a} \right )\left [ \cos ^2h\left ( \dfrac{x}{a} \right ) \right ]^{\frac{1}{2}}$

$=a\cos ^2h\left ( \dfrac{x}{a} \right )$

$=a \times \dfrac{y^2}{a^2}$

$=\dfrac{y^2}{a}$

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 9 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 9 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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