MCQ Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 13

For the curve

$x=t^2-1$ ,

$y=t^2-t$ , the tangent is perpendicular to

$x$ -axis then

0%
$t=0$
0%
$t=\dfrac{1}{2}$
0%
$t=1$
0%
$t=\dfrac{1}{\sqrt{3}}$

Explanation

Given curve is

$x=t^2-1, y=t^2-t$

Derivating w.r.to

$t$ we get

$\dfrac{dx}{dt}=2t$ -------(1)

and

$\dfrac{dy}{dt}=2t-1$ -------(2)

dividing (2) by (1) we get

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

Therefore, the slope of the tangent is

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

Given that the tangent is perpendicular to x-axis. Therefore, tangent is parallel to y-axis.

We know that slope of y-axis is infinity and the slopes of the two parallel lines are equal.

Therefore, slope of the tangent is infinity.

Hence,

$\dfrac{2t-1}{2t}=\dfrac{1}{0}$

$\implies 2t=0 \implies t=0$

If the tangent at any point on the curve

$x^{4} +y^{4}=a^{4}$ cuts off intercepts

$p$ and

$q$ on the coordinate axes the value of

$p^{-4/3}+q^{-4/3}$ is

0%
$a^{-4/3}$
0%
$a^{-1/3}$
0%
$a^{1/2}$
0%
$None\ of\ these$

Explanation

Consider the given expression.

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

Let the point on the curve be $M\left( {{x}_{0}},{{y}_{0}} \right)$ .

Therefore,

$x_{0}^{4}+x_{0}^{4}={{a}^{4}}$

Differentiate the expression given in the question with respect to $x$ .

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

$4{{x}^{3}}+4{{y}^{3}}\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}=-\dfrac{{{x}^{3}}}{{{y}^{3}}}$

So, at the point $M$ , the slope is,

$\Rightarrow -\dfrac{x_{0}^{3}}{y_{0}^{3}}$

Therefore, equation of the tangent is,

$$\begin{align}

$y-{{y}_{0}}=-\dfrac{x_{0}^{3}}{y_{0}^{3}}\left( x-{{x}_{0}} \right)$

$yy_{0}^{3}-y_{0}^{4}=-x_{0}^{3}x+x_{0}^{4}$

\end{align}$$

Let $p$ and $q$ be the $x$ and $y$ intercept, respectively. So, at $x$ intercept,

$-y_{0}^{4}=-x_{0}^{3}p+x_{0}^{4}$

$x_{0}^{3}p=x_{0}^{4}+y_{0}^{4}$

$p=\dfrac{x_{0}^{4}+y_{0}^{4}}{x_{0}^{3}}$

$p=\dfrac{{{a}^{4}}}{x_{0}^{3}}$

Similarly,

$q=\dfrac{{{a}^{4}}}{y_{0}^{3}}$

Now, calculate the value of ${{p}^{-4/3}}+{{q}^{-4/3}}$ .

$={{\left( \dfrac{{{a}^{4}}}{x_{0}^{3}} \right)}^{-4/3}}+{{\left( \dfrac{{{a}^{4}}}{y_{0}^{3}} \right)}^{-4/3}}$

$={{\left( \dfrac{x_{0}^{3}}{{{a}^{4}}} \right)}^{4/3}}+{{\left( \dfrac{y_{0}^{3}}{{{a}^{4}}} \right)}^{4/3}}$

$=\left( \dfrac{x_{0}^{4}}{{{a}^{16/3}}} \right)+\left( \dfrac{y_{0}^{4}}{{{a}^{16/3}}} \right)$

$=\dfrac{x_{0}^{4}+y_{0}^{4}}{{{a}^{16/3}}}$

$=\dfrac{{{a}^{4}}}{{{a}^{16/3}}}$

$={{a}^{-4/3}}$

Hence, this is the required result.

The slope of the tangent to the curve at a point

$(x,y)$ on it is proportional to

$(x-2).$ If the slope of the tangent to the curve at

$(10,-9)$ on it is

$-3$ . The equation of the curves is .

0%
$y=k(x-2)^2$
0%
$y=\dfrac{-3}{16}(x-2)^2+1$
0%
$y=\dfrac{-3}{16}(x-2)^2+3$
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$y=K(x+2)^2$

If the line

$y=4x-5$ touches to the curve

${ y }^{ 2 }=a{ x }^{ 3 }+b$ at the point

$(2,3)$ then

$7a+2b=$

0%
$0$
0%
$1$
0%
$-1$
0%
$2$

Explanation

The slope of given line

$y=4x-5$ is

$m=4$ .

Now, since the given line touches the curve at

$(2,3)$ , the slope of curve

$\dfrac{dy}{dx}$ at the given point is equal to the slope

$m$ of the line.

Hence, for slope of curve, differentiating both sides w.r.t x:-

$2y\dfrac{dy}{dx}=3ax^2$

Putting the value

$x=2$ and

$y=3$

$6\dfrac{dy}{dx}=3a\times 4 \implies 6\times4=12a$

$\implies a=2$

Now, the given point lies on the curve, so it must satisfy the equation of the curve:-

$\implies (3)^2=2\times (2)^3+b$

$\implies b=-7$

$7a+2b=(7\times 2-2\times 7)=0$

Hence, answer is option-(A).

$f(x) = \left\{\begin{matrix} -x^2, & \text{for} \ x < 0 \\ x^2 + 8, & \text{for} \ x \ge 0 \end{matrix}\right.$

Let . Then x-intercept of the line, thet is , the tangent to the graph of f(x) is

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zero
0%
-1
0%
-2
0%
-4

If the tangent at any point on the curve

$x+y^4=a$ cuts off intercepts p and q on the co-ordinate axes, the value of

$p^{-4/3}+q^{-4/3}$ is

0%
$a^{-4/3}$
0%
$a^{-1/2}$
0%
$a^{1/3}$
0%
None of these

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%
${{{x_2}} \over {{x_1}}} + {{{y_2}} \over {{y_1}}} = - 1$
0%
${{{x_2}} \over {{y_1}}} + {{{x_1}} \over {{y_2}}} = - 1$
0%
${{{x_1}} \over {{x_2}}} + {{{y_1}} \over {{y_2}}} = - 1$
0%
${{{x_2}} \over {{x_1}}} + {{{y_2}} \over {{y_1}}} = 1$

Explanation

Given curve equation is

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

The curve passes through the points

$(x_1,y_1)$ and

$(x_2,y_2)$ .

Therefore,

$x_1^2+y_1^2=a^3$ -------(1)

and

$x_2^3+y_2^3=a^3$ -------(2)

subtracting (1) from (2) we get

$y_2^3-y_1^3=-(x_2^3-x_1^3)$ -------(3)

Therefore,

$3x^2+3y^2(\dfrac{dy}{dx})=0$

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

Therefore, the slope of the tangent at

$(x_1,y_1)$ is

$-\dfrac{x_1^2}{y_1^2}$

The tangent passes through the points

$(x_1,y_1)$ and

$(x_2,y_2)$ .

Therefore, the slope of the tangent is

$\dfrac{y_2-y_1}{x_2-x_1}$

On comparing two slopes we get,

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

we know that

$a^3-b^3=(a-b)(a^2+ab+b^2)$

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

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

$\implies x_1^2y_1^2+x_1x_2y_1^2+x_2^2y_1^2=x_1^2y_1^2+x_1^2y_1y_2+x_1^2y_1^2$

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

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

$\implies x-1y_2+x_2y_1=-x_1y_1$

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

A point on the curve

$y = 2{x^3} + 13{x^2} + 5x + 9$ , the tangent at which passes through the origin is

0%
$(1, 15)$
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$(1, -15)$
0%
$(15, 1)$
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$(-1, 15)$

Explanation

Let

$P(a,\,b)$ be a point on the given curve

$y=2x^3+13x^2+5x+9$

$b=2a^3+13a^2+5a+9$ --------- ( 1 )

Now, taking derivative of

$y$ .

$\therefore$

$y'=6x^2+26x+5\Rightarrow y'\,at\,P=6a^2+26a+5$ .

Equation of tangent at

$P$ is

$y-b=(6a^2+26a+5)(x-a)$

This tangent passes through origin means

$b=a(6a^2+26a+5)$

$2a^3+13a^2+5a+9=6a^3+26a^2+5a$

$4a^3+13a^2-9=0$

$(a+1)(a+3)(4a-3)=0$

Taking,

$a+1=0$

We get,

$a=-1$

Substituting

$a=-1$ in equation (1) we get,

$b=2(-1)^3+13(-1)^2+5(-1)+9$

$\therefore$

$b=15$

$\therefore$

$P(a,\,b)=(-1,\,15)$

$\therefore$ A point on the curve

$y=2x^3+13x^2+5x+9$ , the tangent at which passes through the origin is

$(-1,\,15)$ .

The point on the curve

$y = b e^{\dfrac {-x}{a}}$ at which the tangent drawn is

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

0%
$(0, b)$
0%
$\left (a, \dfrac {1}{e}\right )$
0%
$(0, 1)$
0%
$(1, 0)$

The slope of the tangent to the curve

${r^2} = {a^2}\cos 2\theta$ , where

$x = r\cos \theta ,y = r\sin \theta$ , at the point

$\theta=\frac{\pi}{6}$ is

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

The line

$\dfrac{x}{a}+\dfrac{y}{b}=1$ touches the curve

$y=be^{-x/a}$ at the point.

0%
$(a, b/a)$
0%
$(-a, b/a)$
0%
$(0, b)$
0%
None of these

The abscissa of the point on the curve $\sqrt {xy} = a + x$ , the tangent at which cuts off equal intercepts from the co-ordinate axes is (a >0)

0%
$\dfrac{a}{{\sqrt 2 }}$
0%
$- \dfrac{a}{{\sqrt 2 }}$
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$a\sqrt 2$
0%
- $a\sqrt 2$

Explanation

Slope of normal

$= \pm1$ because intercept are equal

slope of target

$=\dfrac{-1}{\left(\dfrac{dy}{dh}\right)}=\pm 1$

$\therefore \dfrac{dy}{dh}=\pm 1$

differentiate by

$xy=(a+x)^2$ we get

$xy^1+y=2(a+x)$

$\Rightarrow y \pm x=2(a+x)$

$\Rightarrow \dfrac{(a+x)^2}{x}\pm x=2(a+x)$

$$\Rightarrow \pm x^2=(a+h)(h-a)$$

$\Rightarrow \pm x^2=h^2-a^2$

$\Rightarrow 2x^2=a^2$ or

$x=\pm \dfrac{a}{\sqrt 2}$

2115529_1108748_ans_2487ee6dd23642f99a2113268967b633.png

$m$ is the slope of a tangent to the curve

$e^{y}=1+x^{2}$ , then

$m$ belongs to the interval

0%
$[-1, 1]$
0%
$[-2, -1]$
0%
$[1, 2]$
0%
$[1, 3]$

Find the slope of tangent of the curve

$x = a\,{\sin ^3}t,y = b\,\,{\cos ^3}t$ at

$t = \frac{\pi }{2}$

0%
$cott$
0%
$-tant$
0%
$-cott$
0%
$\text{not defined at}$ $\frac{\pi}{2}$

Explanation

$x=asin^3t$

$dx=a3sin^2t(cost)dt$ …(1)

$y=bcos^3t$

$dy=b3cos^2t(-sint)dt$ ..(2)

$t=\dfrac{\Pi}{2}$

$\dfrac{dx}{dt}=0$ and hence

$\dfrac{dy}{dx}=\text{not defined}$

The slope of the curve

$y=\sin { x } +\cos ^{ 2 }{ x }$ is zero at a point , whose x-coordinate can be ?

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

A tangent drawn to the curve

$y = f\left( x \right)$ at

$P\left( {x,y} \right)$
cuts the x and y axes at A and B, respectively, such that

$AP:PB = 1:3$ . If

$f\left( 1 \right) = 1$ then the curve passes through

$\left( {k,\frac{1}{8}} \right)$ where

$k$ is

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

Explanation

Equation of tangent at

$P(x,y)$ is

$Y - y = \dfrac{dy}{dt}(X-x)$

It meets x - axis at

$A$

$A = \left(x-\dfrac{dx}{dy}y, 0\right)$

and y -axis at

$B$

$B = (0, y - \dfrac{dy}{dt})$

And also given

$\dfrac{AP}{PB} = 1 : 3$

Using section formula

$\left[\dfrac{mx_1+m_2x_2}{m_1m+m_2}, \dfrac{m_1y_1+m_2y_2}{m_1+m_2}\right]$

$\dfrac{3}{4} \left(x - \dfrac{dx}{dy} y\right)v = \dfrac{1}{4} \left(y - x \dfrac{dy}{dx}\right) = (x,y)$

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

$\dfrac{1}{4} \left(y - x \dfrac{dy}{x} \right) = y$

$3 \dfrac{dx}{x} + \dfrac{dy}{y} = 0$

$\log(x^3y) =$ const (on integrative)

Given

$f(1) = 1$ , so here

$cons \, t = 1$

$x^3y = 1$ (k, 1/8)

$k^3 = 8$

$k = 2$

1331071_1107146_ans_c4d4706b3f1e499a93282a153729ec90.png

Equation of tangent to the circle

$x^{2}+ y^{2}-6x+4y-12=0$ which are parallel to the line

$4x+3y+5=0$ is ?

0%
$4x+3y+19=0$
0%
$4x+3y+31=0$
0%
$4x+3y-19=0$
0%
$4x+3y+29=0$

Explanation

For the circle

$x²+y²-6x+4y-12=0$ ,

center

$C≡\left( \frac { -6 }{ -2 } ,\frac { 4 }{ -2 } \right) =(3,-2)$

radius

$r=\sqrt { (9+4+12) } =5$

A tangent parallel to line

$4x+3y+5=0$ has equation

$4x+3y+c=0\Longrightarrow (1)$

If seg

$CP$ is perpendicular to this tgt,then

$CP=r$

$\left| \frac { 4\left( 3 \right) +3\left( -2 \right) +c }{ \sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 } } } \right| =5$

$\left| c+6 \right| =25$

$c+6=\pm 25$

$c=-6\pm 25$

$c=19,-31\Longrightarrow (2)$

From (1) and (2)

$4x+3y+19=0and4x+3y-31=0.$

Let

$f ( x ) = \frac { 1 } { 1 + x ^ { 2 } }$ . Let

$m$ be the slope,

$'a'$ be the

$x$ -intercept and

$'b'$ be the

$y$ -intercept of tangent to

$y = f ( x )$ .Abscissa of point of contact of the tangent for which

$'m'$ is greatest is:

0%
$\frac { 1 } { \sqrt { 3 } }$
0%
$1$
0%
$0$
0%
$\frac { - 1 } { \sqrt { 3 } }$

$V$ is the set of points on the curve

$y^{3} - 3xy +2 = 0$ where the tangent is vertical then

$V =$ .

0%
$\phi$
0%
$\left \{(1 , 0)\right \}$
0%
$\left \{(1, 1)\right \}$
0%
$\left \{(0, 0), (1, 1)\right \}$

Paraboals

$(y-\alpha )^{2}=4a(x-\beta )and (y-\alpha )^{2}=4a'(x-\beta ')$ will have a common normal (other than the normal passing through vertex ofparabola)if:

0%
$\frac{2(a-a')}{\beta '-\beta }< 1$
0%
$\frac{2(a-a')}{\beta -\beta' }< 1$
0%
$\frac{2(a'-a)}{\beta -\beta' }< 1$
0%
$\frac{2(a'a)}{\beta -\beta' }> 1$

The normal to the curve

$x=a\left( \cos { \theta } +\theta \sin { \theta } \right)$ ,

$y=a\left( \sin { \theta } -\theta \cos { \theta } \right)$ at any point

$\theta$ is such that

0%
it passes through origin
0%
it passes through the point $(1,1)$
0%
it passes through $\left( \frac { a\pi }{ 2 } ,-a \right)$
0%
it is at a constant distance from the origin

Explanation

$x=a\cos \theta +a\theta \sin \theta$

$\dfrac {dx}{d\theta }=-a\sin \theta +a\left\{\theta (-\cos \theta )+\sin \theta \right\}$

$=-a\theta \cos \theta$

$y=a(\sin \theta -\theta \cos \theta )$

$\dfrac {dy}{dx}=a\cos \theta -a \left\{\theta \sin \theta +\cos \theta \right\}$

$\therefore \ \dfrac {dy}{dx}=\dfrac {-a\theta \sin \theta }{-a\theta \cos \theta }=\tan \theta$

The equation of normal at

$\theta$ is:

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

or,

$y-a (\sin \theta -\theta \cos \theta )=-\dfrac {\cos \theta }{\sin \theta }(x-a\cos \theta -a\theta \sin \theta )$

or,

$y \sin \theta -\sin^2\theta +a\theta \sin \theta \cos \theta$

$=-x\cos \theta +a\cos^2\theta +a\theta \sin \theta \cos \theta$

or,

$y\sin \theta +x\cos \theta =a$

which wakes it clear that normal passes through origin

$(0,0)$

The tangent to the circle

$x^2+y^2=5$ at the point

$(1,-2)$ also touches the circle

$x^2+y^2-8x+6y+20=0$ at

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

If for a curve represented parametrically by

$x={ sec }^{ 2 }t,\quad y=cot\quad t\quad$ , the tangent at a point

$P(t=\frac { \pi }{ 4 } )$ meets the curve again at the point Q, then

$\begin{vmatrix} PQ \end{vmatrix}$ is equal to

0%
$\frac { 2\sqrt { 5 } }{ 3 }$
0%
$\frac { 3\sqrt { 5 } }{ 2 }$
0%
$\frac { 5\sqrt { 3 } }{ 3 }$
0%
$\frac { 5\sqrt { 5 } }{ 4 }$

If the subnormal to the curve

${ x }^{ 2 }.{ y }^{ n }={ a }^{ 2 }$ is a constant then n=

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

Explanation

Given,

$x^2y^n=a^2$

$\Rightarrow x=\sqrt{\dfrac{a^2}{y^n}}$

$\Rightarrow 2xy^n+x^2ny^{n-1}\dfrac{dy}{dx}=0$

$\Rightarrow \dfrac{dy}{dx}=-\dfrac{2y}{nx}$

$\Rightarrow \dfrac{dy}{dx}=-\dfrac{2y}{n\sqrt{\frac{a^2}{y^n}}}$

$\therefore \dfrac{dy}{dx}=-\dfrac{2}{na}y^{\frac{n+2}{2}}$

Subnormal is given by,

$SN=y \dfrac{dy}{dx}$

$=y\left ( -\dfrac{2}{na}y^{\frac{n+2}{2}} \right )$

$\therefore SN=-\dfrac{2}{na}y^{\frac{n+4}{2}}$

$\therefore n=-4$

$\theta$ is angle of intersection between

$y=10-x^{2}$ and

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

$|\tan \theta|$ is-

0%
$\dfrac {5\sqrt {3}}{11}$
0%
$\dfrac {7\sqrt {3}}{15}$
0%
$\dfrac {4\sqrt {3}}{11}$
0%
$none$

Explanation

Given:

$y=10-{x}^{2}$ ......

$(1)$

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

$(2)$

On solving,

$10-{x}^{2}=4+{x}^{2}$

$\Rightarrow 2{x}^{2}=6$

$\Rightarrow {x}^{2}=\dfrac{6}{2}=3$

$\therefore x=\pm \sqrt{3}$ from

$(1)$

$\Rightarrow y=10-{\left(\pm\sqrt{3}\right)}^{2}=10-3=7$

Consider

$y=10-{x}^{2}$

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

${m}_{1}=\left[\dfrac{dy}{dx}\right]_{\left(\sqrt{3},7\right)}=-2\times\sqrt{3}=-2\sqrt{3}$

Consider

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

$\dfrac{dy}{dx}=2x$

${m}_{2}=\left[\dfrac{dy}{dx}\right]_{\left(\sqrt{3},7\right)}=2\times\sqrt{3}=2\sqrt{3}$

$=\left|\dfrac{{m}_{1}-{m}_{2}}{1+{m}_{1}{m}_{2}}\right|$

$\tan{\theta}=\left|\dfrac{-2\sqrt{3}-2\sqrt{3}}{1+\left(-2\sqrt{3}\right)\left(2\sqrt{3}\right)}\right|$

$=\left|\dfrac{-4\sqrt{3}}{1-12}\right|$

$=\dfrac{4\sqrt{3}}{11}$

$\therefore \tan{\theta}=\dfrac{4\sqrt{3}}{11}$

The tangent to the curve

$y =x^2 - 5x + 5$ , parallel to the line

$2y = 4x + 1$ , also passes through the point

0%
$\lgroup \dfrac{1}{4}, \dfrac{7}{2} \rgroup$
0%
$\lgroup \dfrac{7}{2}, \dfrac{1}{4} \rgroup$
0%
$\lgroup -\dfrac{1}{8}, 7 \rgroup$
0%
$\lgroup \dfrac{1}{8}, -7 \rgroup$

$y = 6\tan \,x\left( {{e^x} - x - 1} \right) - 3{x^3} - {x^4} - \frac{5}{4}{x^5},\,$ if

${n^{th}}$ derivative at x=0 is non zero then least value of n is

0%
3
0%
4
0%
5
0%
6

The slope of the tangent to the curve at a point (x, y) on it is proportional to (x-2). If the slope of the tangent to the curve at (10,-9) on it -The equation of the curve is

0%
$y=k{ \left( x-2 \right) }^{ 2 }$
0%
$y=\frac { -3 }{ 16 } { \left( x-2 \right) }^{ 2 }\ +1$
0%
$y=\frac { -3 }{ 16 } { \left( x-2 \right) }^{ 2 }\ +3$
0%
$y=k{ \left( x+2 \right) }^{ 2 }$

Explanation

${\textbf{Step -1: Find the value of slope}}{\textbf{.}}$

${\text{The slope of the tangent to the curve at a point }}\left( {x,y} \right){\text{ on it is proportional to }}\left( {x - 2} \right).$

$= \dfrac{{dy}}{{dx}} = m\left( {x - 2} \right)$

$= \dfrac{{dy}}{{dx}} = k\left( {x - 2} \right)\left[ {{\text{where }}k{\text{ is a proportionality constant}}} \right]$

${\text{Slope of the tangent to the curve at }}\left( {10, - 9} \right){\text{ is }} - 3.$

$\Rightarrow - 3 = k\left( {10 - 2} \right)$

$\Rightarrow - 3 = 8k$

$\Rightarrow k = - \dfrac{3}{8}$

$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{3}{8}\left( {x - 2} \right)$

${\textbf{Step -2: Integrate the formed equation and solve it further}}{\textbf{.}}$

${\text{Integrating both sides,}}$

$\Rightarrow y = - \dfrac{3}{8}\dfrac{{{{\left( {x - 2} \right)}^2}}}{2} + c{\text{ }}.......\left( i \right){\text{ }}\left[ {{\text{Eqaution of curve}}} \right]$

${\text{Also, }}\left( {10, - 9} \right){\text{ satisfy the equation of curve as it lies on the curve}}{\text{.}}$

$\therefore - 9 = - \dfrac{3}{8}\dfrac{{{{\left( {10 - 2} \right)}^2}}}{2} + c$

$\Rightarrow - 9 = - \dfrac{3}{{16}} \times {8^2} + c$

$\Rightarrow c = 3$

${\textbf{Hence, the equation of the curve is }}\mathbf{y= - \dfrac{3}{{16}}{\left( {x - 2} \right)^2} + 3.}$

If the curves

$\dfrac {x^{2}}{a^{2}} + \dfrac {y^{2}}{4} = 1$ and

$y^{3} = 16x$ intersect at right angles, then

$a^{2}$ is equal to

0%
$5/3$
0%
$4/3$
0%
$6/11$
0%
$3/2$

Explanation

$y^{3}=16x$

$3y^{2}\dfrac{dy}{dx}=16$

$\dfrac{dy}{dx}=m_{1}=\dfrac{16}{3y^{2}}$

$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{4}=1$

$\dfrac{2x}{a^{2}}+\dfrac{2y}{4}\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}=\dfrac{-x}{a^{2}}\times \dfrac{4}{y}=m_{2}$

$m_{1}.m_{2}=-1$

$\dfrac{16}{3y^{2}}\left(\dfrac{-4x}{a^{2}y}\right)=-1$

$\dfrac{16\times 4\times x}{3a^{2}y^{3}}=1\Rightarrow 64x=3a^{2}y^{3}$

$a^{2}=\dfrac{64 x}{3y^{3}}$ Now

$y^{3}=16 x$

$a^{2}=\dfrac{64 x}{3\times 16 x}\Rightarrow a^{2}=\dfrac{4}{3}$

If the slope of one of the lines given by

${a^2}{x^2} + 2hxy+by^2 = 0$ be three times of the other , then h is equal to

0%
$2\sqrt 3 ab$
0%
$-2\sqrt 3 ab$
0%
$\frac{2}{{\sqrt 3 }}ab$
0%
$-\frac{2}{{\sqrt 3 }}ab$

The equation of normal to the curve

$y=log^x_e$ at the point

$P(1, 0)$ is ___________.

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

Explanation

To find the equation for the normal to the curve

$y = log_e x$ at the point

$P(1,0)$ , we find the slope of said normal and then use the slope-point form of the equation of a line.

We know that the normal and the tangent of a curve at a point on the curve are perpendicular. So,

$\text{Slope of the normal} = \frac{-1}{\text{Slope of the tangent}} = \frac{-1}{y'(1)}$ .

Now,

$y'(x) = \frac{1}{x}$ . So,

$y'(1) = 1$ .

Therefore, the slope of the normal to the curve

$y = log_e x$ at the point

$P(1, 0)$ is

$-1$ .

Noting that, the normal passes through the point

$P$ , we find its equation as:

$y - 0 = -1 \times(x - 1)$ .

That is,

$x + y = 1$ .

If the tangent to the curve

$y=\cfrac { x }{ { x }^{ 2 }-3 } ,x\in R,\left( x\neq \pm \sqrt { 3 } \right)$ , at a point

$\left( \alpha ,\beta \right) \neq \left( 0,0 \right)$ on it is parallel to the line

$2x+6y-11=0$ , then:

0%
$\left| 6\alpha +2\beta \right| =19\quad$
0%
$\left| 2\alpha +6\beta \right| =11$
0%
$\left| 6\alpha +2\beta \right| =9$
0%
$\left| 2\alpha +6\beta \right| =19\quad$

If the line joining the point (0, 3 ) and (5, -2) is a tangent to the curve

$y= \dfrac{c}{x+1}$ , then the value of c is

0%
1
0%
-2
0%
4
0%
None of these

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 13 - 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 13 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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