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

The abscissa of the point on the curve

$3y=6x- 5x^3$ the normal at which passes through origin is :

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$1$
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$\frac {1}{3}$
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$2$
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$\frac {1}{2}$

Explanation

Let

$(x_1,y_1)$ be the point on the given curve

$3y=6x-5x^3$ at which the normal passes through the origin. Then we have

$\left( \frac { dy }{ dx } \right) _{ (x_{ 1 },y_{ 1 }) } =2-5x_1^2 .$

Again the equation of the normal at

$(x_1,y_1)$ passing through the origin gives

$2-5x_1^2= \dfrac{-x_1}{y_1}=\dfrac{-3}{6-5x_1^2}$

Since.

$x_1=1$ satisfies the equation, therefore, correct answer is

$(A) .$

The point of the curve

$y^2 = x$ where the tangent makes an angle of

$\frac { \pi}{4}$ with x-axis is

0%
$( \frac {1}{2}, \frac {1}{4} )$
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$( \frac {1}{4}, \frac {1}{2} )$
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$(4,2 )$
0%
$(1,1)$

Explanation

Given curve is

$y^2 = x$

Therefore as per given condition

$\dfrac{dy}{dx}=\dfrac{1}{2y} = tan \dfrac{\pi}{4}=1 \Rightarrow y=\dfrac{1}{2} \\ \Rightarrow x= \dfrac{1}{4}$

Therefore, correct answer is

$B$ .

The curve

$y=x^{\frac{1}{5}}$ has at

$(0,0)$

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a vertical tangent (parallel to y-axis)
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a horizontal tangent (parallel to x-axis)
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an oblique tangent
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no tangent

Explanation

We have,

$y=x^{1/5}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{5}x^{\tfrac{1}{5}-1}=\frac{1}{5}x^{-4/5}$

$\therefore \left( \dfrac { dy }{ dx } \right) _{ (0,0) }= \tfrac{1}{5} \times (0)^{-4/5}= \infty$

So, the curve

$y=x^{1/5}$ has vertical tangent at

$(0,0)$ , which is parallel to y-axis.

The tangent to the curve

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

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

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

Explanation

The equation of curve is

$y=e^{2x}$

Since, it passes through the point

$(0.1).$

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

$\Rightarrow \left( \frac { dy }{ dx } \right) _{ x=0 }=2e^{2\times0} =2=$ slope of tangent to the curve

$\therefore$ Equation of tangent is

$y-1=2(x-0)$

$\Rightarrow y=2x+1$

Since, tangent to curv

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

$(0,1)$ meets X-axis i.e.,

$y=0$ .

$\therefore 0=2x+1 \Rightarrow x= -\frac{1}{2}$

So, the required point is

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

The equation of tangents to the curve

$y(1+x^2 )=2-x,$ where it crosses x-axis is:

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$x+5y=2$
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$x-5y=2$
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$5x-y=2$
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$5x+y=2$

Explanation

We have, equation of the curve

$y(1+x^2)=2-x...(i)$

$\therefore y.(0+2x)+(1+x^2).\frac{dy}{dx}=0-1$ [on differentiating w.r.t.x]

$\Rightarrow 2xy+ (1+x^2)\frac{dy}{dx}=-1$

$\Rightarrow \dfrac{dy}{dx} = \dfrac{-1-2xy}{1+x^2}...(ii)$

Since, the given curve passes through

$x-$ axis i.e.,

$y=0.$

$\therefore 0(1+x^2)=2-x$ [using Eq.(i)]

$\Rightarrow x=2$

So, the curve passes through the point

$(2,0).$

$\therefore \left( \frac { dy }{ dx } \right) _{ (2,0) }=\dfrac { -1-2\times 0 }{ 1+2^{ 2 } } =-\dfrac { 1 }{ 5 } =$ slope of the curve

$\therefore$ slope of tangent to the curve =

$- \frac{1}{5}$

$\therefore$ Equation of tangent of the curve passing through

$(2,0)$ is

$y-0=-\frac{1}{5}(x-2)$

$\Rightarrow 5y=-x+2$

$\Rightarrow 5y+x=2$

The slope of tangent to the curve

$x=t^2+3t-8,y=2t^2-2t-5$ at the point

$(2,-1)$ is:

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$\frac{22}{7}$
0%
$\frac{6}{7}$
0%
$\frac{-6}{7}$
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$-6$

Explanation

Equation of curve is given is given by

$x=t^2+3t-8$ and

$y=2t^2-2t-5.$

$\therefore \dfrac{dx}{dt}= 2t+3$ and

$\dfrac{dy}{dx}=4t-2$

$\Rightarrow\dfrac{dy}{dx}=\dfrac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } =\dfrac { 4t-2 }{ 2t+3 } ....(i)$

Since, the curve passes through the point

$(2,-1).$

$\therefore 2=t^2+3t-8$

and

$-1=2t^2-2t-5$

$\Rightarrow t^2+3t-10=0$

and

$2t^2-2t-4=0$

$\Rightarrow t^2+5t-2t-10=0$

and

$2t^2+2t-4t-4=0$

$\Rightarrow t(t+5)-2(t+5)=0$

and

$2t(t+1)-4(t+1)=0$

$\Rightarrow (t-2)(t+5)=0$

and

$(2t-4)(t+1)=0$

$\Rightarrow t=2,-5$ and

$t=-1,2$

$\Rightarrow t=2$

$\therefore$ Slope of tangent,

$\left( \frac { dy }{ dx } \right) _{ at\ t-2 }=\dfrac { 4\times 2-2 }{ 2\times 2+3 } =\dfrac { 6 }{ 7 }$ [using Eq.(i)]

The curve for which the slope of the tangent at any point is equal to the ratio of the abscissa to the ordinate of the point is :

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an ellipse
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parabola
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circle
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rectangular hyperbola

Explanation

${\textbf{Step -1: Differentiate slope}}{\textbf{.}}$

${\text{Given: slope of tangent is equal to ratio of abscissa to the ordinate of the point}}{\text{.}}$

${\text{Let slope of the tangent be }}\dfrac{{dy}}{{dx}}.$

$\text{According to the question}$

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

${\textbf{Step -2: Solve further to find the equation}}{\textbf{.}}$

${\text{Separating the variables}}{\text{.}}$

$ydy = xdx$

${\text{Upon integration we get,}}$

$\int {ydy = \int {xdx} }$

$\Rightarrow\dfrac{{{y^2}}}{2} = \dfrac{{{x^2}}}{2} + C$

$\Rightarrow{y^2} = {x^2} + 2C$

$\Rightarrow {x^2} - {y^2} + 2C = 0$

$\Rightarrow {x^2} - {y^2} = k$

$\textbf{[ k = - 2 C]}$

${\textbf{Hence, the given equation is of a rectangular hyperbola}}{\textbf{.}}$

The line

$5x-2y+4k=0$ is tangent to

$4x^{2}-y^{2}=36$ , then k is:

0%
$\dfrac{9}{4}$
0%
$\dfrac{81}{16}$
0%
$\dfrac{4}{9}$
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$\dfrac{2}{3}$

Explanation

The equation of the hyperbola is

$4x^2-y^2=36$

$\dfrac{4x^2}{36}-\dfrac{y^2}{36}=1$

$\dfrac{4x^2}{9}-\dfrac{y^2}{36}=1$ ............(1)

$a^2=9;$

$b^2=36$

The equation of the line is

$5x-2y+4k=0$

$2y=5x+4k$

$y=\dfrac{5}{2}x+2k$ ............(2)

From equation (2)

$m=\dfrac{5}{2},\,c=2k$

$c^2=a^2m^2-b^2$

$(2k)^2=9(\dfrac{5}{2})^2-36$

$4k^2=9\times \dfrac{25}{4}-36$

$4k^2=\dfrac{225-144}{4}$

$k^2=\dfrac{81}{16}\Rightarrow k=\dfrac{9}{4}$

The equation of the tangent to the curve

$y=1-e^{\dfrac{x}{2}}$ at the point of intersection with

$Y-$ axis

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

If the tangent at

$(1,1)$ on

$y^{2}=x(2-x)^{2}$ meets the curve again at

$P$ , then

$P$ is

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$(4,4)$
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$(-1,2)$
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$(3,6)$
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$\left(\dfrac{9}{4}, \dfrac{3}{8}\right)$

Explanation

$\left(\dfrac{9}{4}, \dfrac{3}{8}\right)$
[Hint:

$y^{2}=x(2-x)^{2}=x(4-4x+x^{2})=x^{3}-4x^{2}-4x$

$\therefore 2y\dfrac{dy}{dx}=3x^{2}-8x+4$

$\therefore \dfrac{dy}{dx}=\dfrac{3x^{2}-8x+4}{2y}$

$\therefore \left(\dfrac{dy}{dx}\right)_{at\ (1,1)}=\dfrac{3(1)^{2}-8(1)+4}{2(1)}=\dfrac{1}{2}$
= slope of the tangent at

$(1,1)$

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

$\therefore 2y-12=-x+1$

$\therefore x+2y=3$
Only the coordinates

$\left(\dfrac{9}{4}, \dfrac{3}{8}\right)$ satisfy both the
equation

$y^{2}=x(2-x)^{2}$ and

$x+2y=3$

$\therefore P$ is

$\left(\dfrac{9}{4}, \dfrac{3}{8}\right)$ ].

The slope of the tangent to the curve

$x = t^{2} + 3 t - 8, y = 2t^{2} - 2t - 5$ at the point

$(2, -1)$ is

0%
$\dfrac{22}{7}$
0%
$\dfrac{6}{7}$
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$\dfrac{7}{6}$
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$\dfrac{-6}{7}$

Explanation

$Slope \space of \space tangent \space to \space curve \space y=f(x) \space is \dfrac{dy}{dx}$

$Using\space chain \space rule$

$\dfrac{dy}{dx} = \dfrac{(\dfrac{dy}{dt})}{(\dfrac{dx}{dt})}$

$Given\space x=t^2+3t-8 \implies \dfrac{dx}{dt}=2t+3$

$Given\space y=2t^2-2t-5 \implies \dfrac{dy}{dt}=4t-2$

$\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} = \dfrac{4t - 2}{2t + 3}$

$\dfrac{dy}{dx} |_{t=2} = \dfrac{4 \times 2 - 2}{2 \times 2 + 3} = \dfrac{6}{7}$

$\therefore$ Answer (B)

$\dfrac{6}{7}$

The line

$y = mx + 1$ is a tangent to the curve

$y^{2} = 4x$ if the value of m is .......

0%
$1$
0%
$2$
0%
$3$
0%
$\dfrac{1}{2}$

Explanation

Given,

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

but slope of tangent is

$m = \dfrac{2}{y} \Rightarrow y = \dfrac{2}{m}$

When

$y= \dfrac{2}{m}, x = \dfrac{y^{2}}{4} = 4/m^{2}_{/4} = \dfrac{1}{m^{2}}$

$x = \dfrac{1}{m^{2}}, y = \dfrac{2}{m}$ lie on the line

$y = mx + 1$

$\dfrac{2}{m} = m \times \dfrac{1}{m^{2}} + 1 \Rightarrow \dfrac{2}{m} = \dfrac{1}{m} + 1$

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

Correct answer (A).

The normal at the point

$(1, 1)$ on the curve

$2y + x^{2} - 3$ is .............

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

Explanation

Given,

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

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

Slope of tangent =

$-1$

$\therefore$ equation of normal is

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

$\Rightarrow y - x = 0$

$\Rightarrow x-y = 0$

$\therefore$ Answer (B).)

The intercepts on

$x$ -axis made by tangents to the curve,

$y=\int_{0}^{x}|t|$ dt,

$x\in R$ , which are parallel to the line

$y=2x$ , are equal to

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$\pm 2$
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$\pm 3$
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$\pm 4$
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$\pm 1$

Explanation

$\dfrac{dy}{dx}=|x|=2\Rightarrow x=\pm 2\Rightarrow y=\int_{0}^{2}|t|$ dt

$=2$ for

$x=2$
and

$y=\int_{0}^{-2}|t|$ dt

$=-2$ for

$x=-2$
Hence the equations of the tangents are

$y-2=2(x-2) \Rightarrow y=2x-2$
and

$y+2=2(x+2) \Rightarrow y=2x+2$
Putting

$y=0$ , we get

$x=1$ and

$-1$ .

The normal to the curve

$x^{2} = 4y$ passing

$(1, 2)$ is

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

Explanation

Given,

$4y = x^{2} \Rightarrow 4 \dfrac{dy}{dx} = 2x \Rightarrow \dfrac{dy}{dx}|_{x = x1} = \dfrac{x_{1}}{2}$

slope of tangent is

$= \dfrac{x_{1}}{2}$

slope of normal is

$= \dfrac{-2}{x_{1}}$

equation of normal

$y - y_{1} = \dfrac{-2}{x_{1}} (x - x_{1})$

but this passes they

$(1, 2)$

$2 - y_{1} = \dfrac{-2}{x} (1 - x) \Rightarrow 2 - y_{1} = \dfrac{-2}{x_{1}} + 2$

$y_{1} = \dfrac{2}{x_{1}}$

but

$x^{2} = 4y$

$x^{2}_{1} = 4y_{1}$

$x^{2}_{1} = 4 \times \dfrac{2}{x_{1}} = 8 \Rightarrow x_{1} = 2$

when

$x_{1} = 2 y_{1} = \dfrac{2}{2} = 1$

$\therefore$ equation of normal is

$y - 1 = \dfrac{-2}{2} \times (x - 2) \Rightarrow y - 1 = (x - 2)$

$x + y = 3$

The slope of the normal to the curve

$y = 2x ^{2} + 3 \sin x$ at

$x = 0$ is

0%
$3$
0%
$1/3$
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$-3$
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$-1 /3$

Explanation

Given,

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

slope

$= \dfrac{dy}{dx} = 4x + 3 \cos x$

$\dfrac{dy}{dy}|_{x=0} = 4 ( 0 ) = 3$

slope of normal =

$-1/ \dfrac{dy}{dx} = -1/3$

The line

$y = x + 1$ is a tangent to the curve

$y^{2} = 4 x$ at the point

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

Explanation

Given,

$y^{2} = 4 x \\ \Rightarrow 2y \dfrac{dx}{dx} = 4 \\ \Rightarrow \dfrac{dx}{dy} = 2/y$

but given that

$y = x + 1$ is a tangent to the curve its slope =

$1 ( y = mx +c )$

$\dfrac 2y =1\implies y=2$

$x=2-1=1$

The line

$y = x + 1$ is a tangent to the curve

$y^{2} = 4 x$ at the point

$(1,2)$ .

The points on the curve

$9 y^{2} = x^{3}$ , where the normal to the curve makes equal intercepts with the axes are ...........

0%
$\left ( 4, \pm \dfrac{8}{3} \right )$
0%
$\left ( 4, \dfrac{-8}{3} \right )$
0%
$\left ( 4, + \dfrac{8}{3} \right )$
0%
$\left (\pm 4, \dfrac{8}{3} \right )$

The point of contact of vertical tangent to the curve given by the equations

$\mathrm{x}=3-2\cos\theta, \mathrm{y}=2+3\sin\theta$ is

0%
(1, 5)
0%
(1, 2)
0%
(5, 2)
0%
(2, 5)

Explanation

$x=3-2cos\theta ,y=2+3sin\, \theta$
The slope of the tangent at

$'\theta '$ is

$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{d\theta }}{\dfrac{dx}{d\theta }}$

$\therefore$ for vertical tangent

$\dfrac{dx}{d\theta }=0i.e.,2sin\, \theta =0$

$\Rightarrow \theta =0\,$ or

$\pi or2\pi$
Then

$x=1,y=2;x=5 ,y=2;x=1,y=2$

$\therefore$ the points of contact are

$(1,2) ,(5,2)$

The coordinates of a point P(x, y) lying in the first quadrant of the ellipse

$\displaystyle \frac{x^{2}}{8}+\frac{y^{2}}{18}=1$ so that the area of the triangle formed by the tangent at P and the axes is the smallest are

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

Explanation

Let point be

$(2\sqrt 2 sin\theta, 3\sqrt 2 cos\theta)$

$y'=\dfrac {-3\sqrt 2}{2\sqrt 2}tan\theta$

$y'=\dfrac {-3}{2}tan\theta$

$(y-3\sqrt 2cos\theta)=\dfrac {-3}{2}tan\theta (x-2\sqrt 2 sin\theta)$

$x=\dfrac {2}{sin\theta}$ (x intercept)

$y=\dfrac {3}{cos\theta}$ (y intercept)

so area

$=\dfrac {1}{2}\dfrac {\times 2\times 3}{sin\theta cos\theta}$

area is min when

$sin 2\theta=1$

$\theta=\dfrac {\pi}{4}$

so point (2,3)

The greatest inclination between the tangents is

0%
$\displaystyle \tan^{-1} \left ( \dfrac{\mathrm{a}+\mathrm{b}}{2\sqrt{\mathrm{a}\mathrm{b}}}\right )$
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$\displaystyle \tan^{-1} \left (\dfrac{\mathrm{a}-\mathrm{b}}{2\sqrt{\mathrm{a}\mathrm{b}}}\right )$
0%
$\tan^{-1}\sqrt{\dfrac{\mathrm{a}}{\mathrm{b}}}$
0%
$\tan^{-1}\sqrt{\dfrac{\mathrm{b}}{\mathrm{a}}}$

Explanation

The slope of the tangent to the ellipse

$=-\displaystyle \dfrac{\mathrm{b}}{\mathrm{a}}\cot\theta$
The slope of tangent to the circle is

$-\cot\theta$

$\therefore$ the angle

$\phi$ between the tangents is given by

$\tan\phi=\dfrac{(-\dfrac{\mathrm{b}}{\mathrm{a}}+1)\cot\theta}{1+\dfrac{\mathrm{b}}{\mathrm{a}}\cot^{2}\theta}=\dfrac{(\mathrm{a}-\mathrm{b})}{\mathrm{a}\tan\theta+\mathrm{b}\cot\theta}=\dfrac{\mathrm{a}-\mathrm{b}}{(\sqrt{\mathrm{a}\tan\theta}-\sqrt{\mathrm{b}\cot\theta})^{2}+2\sqrt{\mathrm{a}\mathrm{b}}}$
This is maximum when

$\sqrt{\mathrm{a}\tan\theta}-\sqrt{\mathrm{b}\cot\theta}=0$

$\therefore$ the maximum angle

$=\displaystyle \tan^{-1}(\dfrac{\mathrm{a}-\mathrm{b}}{2\sqrt{\mathrm{a}\mathrm{b}}})$

$OT$ and

$ON$ are perpendiculars dropped from the origin to the tangent and normal to the curve

$x=a\sin^{3}t, y=a\cos^{3}t$ at an arbitrary point, then which of the following is/are correct?

0%
$4{OT}^{2}+ON{}^{2}=a^{2}$
0%
${OT}^{2}+{ON}^{2}=a^{2}$
0%
${OT}^{2}+{ON}^{2}=2a^{2}$
0%
${OT}^{2}+2{ON}^{2}=4a^{2}$

Explanation

Given equation of curve is

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

$\displaystyle \frac{dx}{dt}=3a\sin^2 t\cos t$

$\displaystyle \frac{dy}{dt}=-3a\cos^2 t\sin t$

$\Rightarrow \displaystyle \frac{dy}{dx}=-\cot t$

Slope of tangent to curve at

$t$ is

$-\cot t$

Equation of tangent at

$t$ is

$y-a\cos^ 3 t =-\cot t(x-a\sin^3 t)$

$\Rightarrow -x\cot t -y +a\cos^3 t+a\cos t \sin^2 t =0$

Now,

$OT= \left|\cfrac{a\cos^3 t+a\cos t\sin^2 t}{\sqrt{1+\cot^2 t }}\right|$

$\Rightarrow OT=|a\cos t \sin^3 t +a\cos^3 t \sin t|$

$\Rightarrow OT^2 =a^2\sin ^2t \cos^2 t (\sin^2 t+\cos ^2 t )^2$

$\Rightarrow 4OT^2=4a^2 \sin^2 t \cos ^2 t$ ......(1)

Now, slope of normal to curve at

$t$ is

$\tan t$

Equation of normal at

$t$ is

$y-a\cos^ 3 t =\tan t(x-a\sin^3 t)$

$\Rightarrow x\tan t -y +a\cos^3 t-a\tan t \sin^3 t =0$

Now,

$ON= \left|\cfrac{a\cos^3 t-a\tan t\sin^3 t}{\sqrt{1+\tan^2 t }}\right|$

$\Rightarrow ON=|a\cos^4 t - a\sin^4 t |$

$\Rightarrow ON^2 =a^2 (\cos^4 t- \sin^4 t )^2$

$\Rightarrow ON^2=a^2 (\cos ^2 t-\sin^ 2t)^2$

$\Rightarrow ON^2=a^2(\cos^2 t+\sin^2t)^2-4a^2\sin^2t \cos ^2t$

$\Rightarrow ON^2=a^2-4a^2\sin^2 t\cos^2 t$ .....(2)

Adding (1) and (2), we get

$4OT^2+ON^2=a^2$

Hence, option A.

A function

$y=f(x)$ has a second order derivative

$f''(x)=6(x-1)$ .
If its graph passes through the point

$(2,1)$ and at that point the tangent to the graph is

$y=3x-5$ , then the function is

0%
$(x-1)^{2}$
0%
$(x+1)^{2}$
0%
$(x+1)^{3}$
0%
$(x-1)^{3}$

Explanation

$f''(x)=6(x-1)$

$\Rightarrow f'(x) = 3x^2 -6x+k$ ...(1)
Slope of tangent at

$(2,1)$ is

$3$ .

$\Rightarrow 3=3 \times 2^2 -6 \times 2 +k$

$\Rightarrow k=3$ ...(2)
From (1) and (2)

$f(x)=x^3-3x^2+3x+c$
Using

$f(x)=y=1$ when

$x=2$ , we get

$1=2+c \Rightarrow c=-1$
Hence, the function is

$f(x)=x^3-3x^2+3x-1$

$\Rightarrow f(x)=(x-1)^3$

The portion of the tangent to the curve

$x=\displaystyle \sqrt{a^{2}-y^{2}}+\frac{a}{2}\log\dfrac{a-\sqrt{a^2-y^2}}{a+\sqrt{a^{2}-y^{2}}}$ intercepted between the curve and

$x-$ axis, is of length

0%
$\displaystyle \frac{a}{2}$
0%
$a$
0%
$2 a$
0%
$\displaystyle \frac{a}{4}$

Explanation

$x=\displaystyle \sqrt{a^{2}-y^{2}}+\frac{a}{2}\log\dfrac{a-\sqrt{a^2-y^2}}{a+\sqrt{a^{2}-y^{2}}}$ .....(1)

Differentiating w.r.t.

$x$ , we get

$1=\displaystyle \frac { -y }{ \sqrt { a^{ 2 }-y^{ 2 } } } \frac { dy }{ dx } +\frac { ay }{ 2(a-\sqrt { a^{ 2 }-y^{ 2 } } )\sqrt { a^{ 2 }-y^{ 2 } } } \frac { dy }{ dx } +\frac { ay }{ 2(a+\sqrt { a^{ 2 }-y^{ 2 } } )\sqrt { a^{ 2 }-y^{ 2 } } } \frac { dy }{ dx }$

$\Rightarrow \displaystyle \frac { 2\sqrt { a^{ 2 }-y^{ 2 } } }{ y } =\left( -2+\frac { a }{ (a-\sqrt { a^{ 2 }-y^{ 2 } } ) } +\frac { a }{ (a+\sqrt { a^{ 2 }-y^{ 2 } } ) } \right) \frac { dy }{ dx }$

$\Rightarrow \displaystyle \frac { 2\sqrt { a^{ 2 }-y^{ 2 } } }{ y } =2\left( \frac { a^{ 2 }-y^{ 2 } }{ y^{ 2 } } \right) \frac { dy }{ dx }$

$\Rightarrow \displaystyle \frac{dy}{dx}=\cfrac{y}{\sqrt{a^2-y^2}}$

Slope of tangent at

$P(x_1,y_1)$ is

$\cfrac { { y }_{ 1 } }{ \sqrt { a^{ 2 }-{ y }_{ 1 }^{ 2 } } }$

Equation of tangent at

$P(x_1,y_1)$ is

$y-y_{ 1 }=\cfrac { { y }_{ 1 } }{ \sqrt { a^{ 2 }-{ y }_{ 1 }^{ 2 } } } (x-x_{ 1 })$ ....(2)

Substituting

$y=0$ in eqn (1), we get

$x=a$

So, the point A on axis is

$(a,0)$

Since, the tangent (given by eqn (2)) passes through

$(a,0)$

$\Rightarrow (x_1-a)=\sqrt{a^2-y_1^2}$ .....(3)

Now, portion of tangent between curve and x-axis is

$AP=\sqrt{(x_1-a)^2+y_1^2}$

$\Rightarrow AP=\sqrt{a^2}$ (by (3))

$\Rightarrow AP=a$

Hence, option B.

A curve passes through

$(2, 0)$ and the slope of the tangent at any point

$(x, y)$ is

$x^2 -2x$ for all values of

$x$ . The point of minimum ordinate on the curve where

$x > 0$ is

$(a, b)$ '

Then find the value of

$a + 6b$ .

0%
$2$
0%
$4$
0%
$-2$
0%
$-4$

Explanation

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

$y = \dfrac {x^3}{3} - x^2 + C$
Passing through (2, 0)

$\Rightarrow C = \dfrac {4}{3}$

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

$y' = x^2 - 2x = x(x - 2)$
For maxima or minima,

$y'=0$

$\Rightarrow x= 0,2$

$y''>0$ at

$x=2$
At x = 2, y takes the minimum value.

$\therefore$ minimum value of y is

$= \dfrac {8}{3} - 4 +\dfrac {4}{3} = 0$

$\therefore a = 2, b = 0$

$a + 6b = 2$

The value of

$x$ at which tangent to the curve

$y=x^3-6x^2+9x+4, 0\leq x \leq 5$ has maximum slope is

0%
$0$
0%
$2$
0%
$\dfrac{5}{2}$
0%
$5$

Explanation

$y=x^3-6x^2+9x+4$

$\dfrac{dy}{dx}=3x^2-12x+9$

Slope of tangent,

$m=f(x)=3x^2-12x+9$

For maxima or minima,

$f'(x)= 0$

$\Rightarrow 6x-12=0$

$\Rightarrow x=2$ .

Now,

$m(0)=9, m(2)=-3$ and

$m(5)=24$
Hence,

The maximum value is attained at

$x = 5$ .

Hence, option D.

The number of different points on the curve

$y^2=x(x+1)^2$ , where the tangent to the curve drawn at (1, 2) meets the curve again, is

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

Explanation

Differentiating with respect to y we get

$y'=\displaystyle \frac{3x^3+4x+1}{2y}$

Therefore for the slope of the tangent passing through (1,2)

$y' =$

$m =$

$\displaystyle \frac{3+4+1}{2(2)}$

$=2$

Hence equation of tangent line is

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

$y=2x$
Therefore for the point of intersection

$(2x)^{2}=x(x+1)^{2}$

$x(x-1)^{2}=0$

$x=0$ and

$x=1$
Substituting in the equation of the tangent,

$x=0 \Rightarrow y=0$

$x =1 \Rightarrow y=2$ ,

But

$(1,2)$ , is the point at which the tangent is drawn. Hence the tangent meets the curve at one more point.

Suppose

$a,b,c$ are such that the curve

$y = ax^2 + bx + c$ is tangent to

$y = 3x -3$ at

$(1, 0)$ and is also tangent to

$y = x + 1$ at

$(3, 4)$ then the value of

$(2a -b -4c)$ equals

0%
$7$
0%
$8$
0%
$9$
0%
$10$

Explanation

$y = ax^2 + bx + c$
Since

$(1,0)$ lies on the curve

$\Rightarrow a + b + c = 0$ ...... (i)
Slope of curve

$\displaystyle =\frac {dy}{dx} = 2ax + b$
Slope of curve at (1,0)

$=2a+b$
Slope of curve at (3,4)

$=6a+b$
Slope of tangent

$y=3x-3$ is 3
Slope of tangent

$y=x+1$ is 1

$(\dfrac {dy}{dx})_{x=1} = 3$ and

$(\dfrac {dy}{dx})_{x=3} = 1$

$2a + b = 3$ .... (ii)

$6a + b = 1$ .... (iii)
on solving we get a

$= -\dfrac {1}{2}, b = 4, c = -\dfrac {7}{2}$

$\therefore 2a - b - 4c = -1 -4 + 14 = 9$

The number of values of c such that the straight line

$3x + 4y = c$ touches the curve

$\dfrac{x^{4}}{2}=x+y$ , is :

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

Explanation

Equation of curve is

$x^4=2x+2y$

$\displaystyle \frac{dy}{dx}=2x^{3}-1$
Equation of line is

$3x+4y=c$
Slope of line is

$-\dfrac{3}{4}$
Since, the line touches the curve,
Slope of curve

$=$ Slope of tangent

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

$\Rightarrow x=\dfrac{1}{2}$

$\Rightarrow y=-\dfrac{15}{32}$

$\therefore$ only one value of c is possible

Let tangent at a point P on the curve

$x^{2m}\: Y^{\frac{n}{2}}=a^{\frac{4m+ n}{2}}(m, \: n\in \: N, \: n \: is \: even)$ , meets the x-axis and y-axis at A and B respectively, if

$AP : PB \:is \:\dfrac{n}{\lambda m}$ , where P lies between A and B, then find the value of

$\lambda$

0%
2
0%
4
0%
6
0%
8

Explanation

Given curve is

$x^{2m}\: y^{\displaystyle\frac{n}{2}}=a^{\displaystyle\frac{4m+ n}{2}}$
applying log on both sides gives

$\displaystyle 2m\log {x}+\frac{n}{2}\log {y}=\frac{4m+n}{2}\log {a}$
differentiating w.r.t

$x$

$\displaystyle \frac{2m}{x}+\frac{n}{2y}\frac{dy}{dx}=0$

$\Rightarrow \displaystyle \frac{dy}{dx}=-\frac{4m}{n}\frac{y}{x}$
Equation of tangent at

$P(x_{1},y_{1})$ is

$\displaystyle y-y_{1}=-\frac{4m}{n}\frac{y_{1}}{x_{1}}\left(x-x_{1}\right)$
above equation meets

$x$ -axis at

$A$ and

$y$ -axis at

$B$ respectively

$A\left(\displaystyle\frac {4m+n}{4m}x_{1},0\right)$

$B\left(0,\displaystyle\frac{4m+n}{n}y_{1}\right)$
given

$AP:PB$ =

$\displaystyle\frac{n}{\lambda m}$

$\Rightarrow \displaystyle\frac{\lambda m\left(\displaystyle\frac{4m+n}{4m}\right)x_{1}}{\lambda m+n}=x_{1}$ (by using section formula)

$\therefore$

$\lambda=4$
Hence, option B.

The equation of the normal to the curve

$y = e^{-2|x|}$ at the point where the curve cuts the line

$x=\displaystyle \frac{1}{2}$ is

0%
$2e(ex + 2y) = e^{2} - 4$
0%
$2e(ex - 2y) = e^{2} - 4$
0%
$2e(ey - 2x) = e^{2} - 4$
0%
none of these

Explanation

Given equation of curve is

$y=e^{-2\left | x \right |}$
At

$x = \dfrac{1}{2}\Rightarrow y=e^{-1}$

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

$\left.\begin{matrix} \dfrac{dy}{dx} \end{matrix}\right|_{x=\frac{1}{2}}=-2e^{-1}$
Slope of tangent at

$\displaystyle (\frac{1}{2}, e^{-1})=\frac{-2}{e}$
Slope of normal

$\displaystyle =\frac{e}{2}$
Equation of normal at

$\displaystyle (\frac{1}{2}, e^{-1})$ is

$\displaystyle y-\frac{1}{e}=\frac{e}{2}\left ( x-\frac{1}{2} \right )$

$\displaystyle y-\frac{ex}{2}=\frac{1}{e}-\frac{e}{4}$

$\displaystyle\frac{2y-ex}{2}=\frac{4-e^{2}}{4e}$

$\Rightarrow2e(ex- 2y) = e^{2}-4$

The perpendicular distance between the point (1, 1) and the tangent to the curve y

$=e^{2x}+x^2$ drawn at the point x

$=$ 0 is

0%
$\dfrac{1}{\sqrt{5}}$
0%
$\dfrac{3}{\sqrt{5}}$
0%
$\dfrac{2}{\sqrt{5}}$
0%
$\dfrac{4}{\sqrt{5}}$

Explanation

The equation of given curve is

$y = e^{2x} + x^2$ ........(i)

$x = 0 \Rightarrow y = 1$

$\therefore$ the given point is P(0, 1)

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

Slope of tangent at (0,1) is

$\displaystyle (\dfrac{dy}{dx})_{(0,1)}=2$

Equation of tangent at P(0, 1) to

$eq^n(i)$ is

$y - 1 = 2(x - 0)$

$\Rightarrow 2x - y + 1 = 0$

$\therefore$ Required distance

$=$ length of perpendicular from (1, 1) to the line

$2x -y + 1 = 0$

$=|\dfrac{2-1+1}{\sqrt{4+1}}|=\dfrac{2}{\sqrt{5}}$

The sum of the intercepts on the coordinate axis by any tangent to the curve

$\sqrt{x} + \sqrt{y} = 2$ is

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

Let

$x=4 \cos ^4{\theta}$ and

$y=4\sin ^4 \theta$

$\Rightarrow \dfrac{dy}{dx}=-\tan^2\theta$

Equation of the tangent at

$(4 \cos ^4 \theta, 4 \sin ^4 \theta)$ is

$\displaystyle\frac{y-4 \sin ^4 \theta}{x-4 \cos ^4 \theta}=-\tan^2 \theta$

$\Rightarrow x$ intercept

$= 4 \cos ^2 \theta$

$\Rightarrow y$ intercept

$= 4 \sin ^2\theta$

So, the sum of intercept made on the coordinate axes

$=4 (\sin ^2 \theta + \cos ^2 \theta)=4$

For the curve represented parametrically by the equations,

$x = 2 ln \cot( t) + 1$ &

$y = \tan( t) + \cot( t)$

0%
tangent at $t = \pi/4$ is parallel to x - axis
0%
normal at $t = \pi/4$ is parallel to y - axis
0%
tangent at $t = \pi/4$ is parallel to the line $y = x$
0%
tangent and normal intersect at the point $(2, 1)$

Explanation

$x=2ln cot(t)+1$

$\displaystyle \frac{dx}{dt}=\frac{2}{cot(t)}(-cosec^2 t)=\frac{-2}{sin^2( t) {cot}(t)}$

$\displaystyle \left(\frac{dx}{dt}\right)_{ t=\frac{\pi}{4}} =-4$

$y=tan(t) +cot(t)$

$\displaystyle \frac{dy}{dt}=sec^2 t-cosec^2t=\frac{sin^2t-cos^2t}{sin^2t cos^2t}$

$\displaystyle \left (\dfrac{dy}{dt}\right )_{ t=\frac{\pi}{4} }=0$

$\Rightarrow \dfrac{dy}{dx}=0$

Slope of tangent

$=0$

$\Rightarrow$ Tangent is parallel to x-axis.

Slope of normal

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

$\Rightarrow$ Normal is parallel to y-axis.

The equation of the tangent to the curve

$y=\sqrt{9-2x^2}$ at the point, where the ordinate & the abscissa are equal , is

0%
$2x+y-\sqrt3=0$
0%
$2x+y-3=0$
0%
$2x-y-3\sqrt3=0$
0%
$2x+y-3\sqrt{3}=0$

Explanation

Let

$(x_{1}, y_{1})$ be any point on the curve such that

$x_1=y_1$

$\therefore {x_{1}}^{2}={9-2x_{1}^{2}}$

$\Rightarrow x_{1}=\pm{\sqrt{3}}$

$\Rightarrow y_{1}=\pm{\sqrt{3}}$

Since,

$y > 0$ , therefore the point is

$(\sqrt3,\sqrt3)$

We have

$y^2=9-2x^2$

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

$\therefore \left ( \dfrac{dy}{dx} \right )_{(\sqrt{3},\sqrt{3})}=-2$

So, the equation of tangent is

$(y-\sqrt{3})=-2(x-\sqrt{3})$

$\Rightarrow 2x+y-3\sqrt3=0$

The point on the curve

$y^{2} = x ,$ the tangent at which makes an angle of

$45^{0}$ with positive direction of

$x -$ axis will be given by

0%
$\left (\displaystyle \frac{1}{2},\displaystyle \frac{1}{4} \right )$
0%
$\left ( \displaystyle \frac{1}{2}, \displaystyle \frac{1}{2} \right )$
0%
$(2,4)$
0%
$\left ( \displaystyle \frac{1}{4}, \displaystyle \frac{1}{2} \right )$

Explanation

$\textbf{Step 1: Find the slope of tangent to given curve at any point}$

$\text{ Given equation of curve is } y^{2} = x$ ...(1)

$\text{Differentiating w.r.t. x we get }$

$\Rightarrow 2y\dfrac{dy}{dx}=1$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2y}$

$\text{Hence slope of tangent to given curve at any point } (x, y) \text{ is }\dfrac{dy}{dx}=\dfrac{1}{2y}$

$\textbf{Step 2: Find the required point}$

$\text{Given slope of tangent}= \tan 45^{\circ}$

$\text{Hence }\displaystyle\frac{1}{2y}=1\Rightarrow y=\frac{1}{2}$

$\displaystyle \Rightarrow x=\frac{1}{4}$ [From eq.(1)]

$\text{Hence, the required point is } \left ( \dfrac{1}{4},\dfrac{1}{2} \right )$

$\textbf{Hence, Option 'D' is correct}$

At the point

$P(a,a'')$ on the graph of

$y=x^n$ ,

$(n \epsilon N)$ , in the first quadrant , a normal is drawn. The normal intersects the

$y$ -axis at the point

$(0,b)$ . If

$\lim_{a\rightarrow 0}b=\displaystyle \frac{1}{2}$ , then n equals

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

Explanation

$y=x^{n}$

$\displaystyle \dfrac{dy}{dx}=nx^{n-1}$

$\displaystyle (\dfrac{dy}{dx})_{(a,a^{n})}=na^{n-1}$

Slope of normal at P,

$=-\dfrac{1}{na^{n-1}}$

Equation of normal at P is

$y-{ a }^{ n }=-\dfrac { 1 }{ n{ a }^{ n-1 } } (x-a)$

Since, it intersects y-axis at (0,b)

$\Rightarrow b-a^{n}=\dfrac{a}{na^{n-1}}$

$\Rightarrow \lim _{ a\rightarrow 0 }{ b- } \lim _{ a\rightarrow 0 }{ { a }^{ n } } =\lim _{ a\rightarrow 0 }{ (\dfrac { a }{ n{ a }^{ n-1 } } ) }$

$\Rightarrow \dfrac { 1 }{ 2 } =\lim _{ a\rightarrow 0 }{ (\dfrac { a }{ n{ a }^{ n-1 } } ) }$ ......(i)

Now, we will check by options.

Option C, satisfies (i)

Hence,

$n=2$

Let

$f(x)$ =

$\begin{cases} -x^2, {for \ x<0} \\x^2+8, {for \ x\geq 0} \end{cases}$ . Then

$x$ -intercept of the line, that is, the tangent to the graph of

$f(x)$ in both the intervals of its domain, is

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

Explanation

At any point $(k, k^2+8)$ on $y=x^2+8$ , the slope of tangent is $\dfrac{dy}{dx}|_{x=k}=2k$ .....(1)

Similarly, at any point $(p, -p^2)$ on $y=-x^2$ , the slope of tangent is $\dfrac{dy}{dx}|_{x=p}=-2p$ ....(2)

For common tangent, $2k=-2p \Rightarrow k=-p$

Hence, the coordinates of point on common tangent and on $y=-x^2$ can be written as $(-k, -k^2)$ as shown in figure.

So, slope of common tangent is $\dfrac{8+k^2-(-k^2)}{k-(-k)}=\dfrac{8+2k^2}{2k}$

From (1), slope is also equal to $2k$ .

Hence, $\dfrac{8+2k^2}{2k}=2k$

Solving, we get $k= \pm 2$ .

Discarding negative value (since 'k' is positive as defined in the function), we get $k=2$

Hence, slope of common tangent is $2k=4$

The point common to tangent and $y=x^2+8$ is $(2,12)$

Hence, equation of common tangent is $y-12=4(x-2)$

To find $x$ intercept, substitute $y=0$ in the above equation.

We get $x$ intercept $=-1$

A function

$y=f(x)$ has a second-order derivative

$f''(x)=6(x-1)$ . If its graph passes through the point

$(2,1)$ and at the point tangent to the graph is

$y=3x-5$ , then the value of

$f(0)$ is

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

Explanation

$f''(x)=6x-6$

$\int { f''(x)dx } =\int { (6x-6)dx }$

$\Rightarrow f'(x)=3x^{2}-6x+C$
Since, it passes through (2,1) and tangent is

$y=3x-5$

$\Rightarrow C=3$
So,

$f'(x)=3x^{2}-6x+3$

$\int { f'(x)dx } =\int { (3{ x }^{ 2 }-6x+3)dx }$

$\Rightarrow f(x)=x^{3}-3x^{2}+3x+C_1$
Since, it passes through

$(2,1)$

$\Rightarrow C_1=-1$
Hence,

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

$f(0)=-1$

For the curve

$y=3 \sin \theta \cos \theta, x= e^{\theta} \sin \theta, 0 \leq \theta \leq \pi$ , the tangent is parallel to x-axis when

$\theta$ is :

0%
$\displaystyle \frac{\pi}{4}$
0%
$\displaystyle \frac{\pi}{2}$
0%
$\displaystyle \frac{3\pi}{4}$
0%
$\displaystyle \frac{\pi}{6}$

Explanation

$y=3\sin { \theta } \cos { \theta }$

$\Rightarrow \displaystyle\frac { dy }{ d\theta } =3(\cos { \theta } -\sin { \theta } )(\cos { \theta } +\sin { \theta } )$

$x={ e }^{ \theta }\sin { \theta }$

$\displaystyle\frac { dx }{ d\theta } =3{ e }^{ \theta }(\cos { \theta } -\sin { \theta } )$

$\displaystyle\frac { dy }{ dx} = 3{ e }^{ -\theta }(\cos { \theta } -\sin { \theta })$

$\displaystyle\frac { dy }{ dx} =0$

$\Rightarrow \sin { \theta } =\cos { \theta }$

$\Rightarrow \displaystyle \theta = \frac{\pi}{4}$

If the circle

$x^2+y^2+2gx+2fy+c=0$ is touched by

$y=x$ at P such that OP =

$6\sqrt{2}$
then the value of c is

0%
36
0%
144
0%
72
0%
None of these

Explanation

Let the point of contact be

$x_{1},y_{1}$
However it lies on the line

$y=x$
Hence

$x_{1}=y_{1}$
Applying distance formula, we get

$\sqrt{x_{1}^2+x_{1}^2}=6\sqrt{2}$

$2x_{1}^2=72$

$x_{1}=6$ ...(positive, since it lies on y=x.)
Differentiating the equation of circle with respect to x.

$2x+2yy'+2g+2fy'=0$
Now

$y'=1$ since

$y'$ is the slope of the line

$y=x$ .
By substituting, we get

$x+y=-(g+f)$
Now

$x_{1}=y_{1}=6$
Hence

$12=-(g+f)$ ...(i)
Substituting

$x=y=6$ in the equation of the circle, we get

$36+36+2(6)(g+f)+c=0$

$72+12(-12)+c=0$

$c=144-72$

$c=72$

The tangent of the acute angle between the curves

$y=|x^2-1|$ and

$y=\sqrt {7-x^2}$ at their points of intersection is

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

Explanation

The given curves are

$y=|x^{2}-1|$ and

$y=\sqrt{7-x^2}$

To find the point of intersection, let us equate both the curves.

$|x^{2}-1|=\sqrt{7-x^2}$

Squaring on both sides gives us

$x^{4}-x^{2}-6=0$

$(x^2-3)(x^2+2) =0$

On solving above quadratic equation, we get

$x^{2}=3$

The point of intersection is

$\left(\pm \sqrt{3},2\right)$

The slope of the tangent to

$y=|x^{2}-1|$ at the point

$\left( \sqrt{3},2\right)$ is

$m_{1}=\displaystyle\dfrac{dy}{dx}=2x=2\sqrt{3}$

The slope of tangent to

$y=\sqrt{7-x^2}$ at the point

$\left(\sqrt{3},2\right)$ is

$m_{2}=\displaystyle\dfrac{dy}{dx}=\dfrac{-x}{\sqrt{7-x^2}}=\dfrac{-\sqrt{3}}{2}$

Let

$\theta$ be the acute angle between the two tangents.

$\displaystyle\tan\theta = \dfrac{m_{1}-m_{2}}{1+m_{1}m_{2}}$

$\Rightarrow \displaystyle\tan\theta = \dfrac{-5\sqrt{3}}{4}$

$\therefore$ The tangent of the acute angle between the curves is

$\displaystyle\dfrac{5\sqrt{3}}{4}$ .

Hence, option C is correct.

The angle made by the tangent of the curve

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

$y=a(1+sint)^2$ with the

$x- axis$ at any point on it is

0%
$\displaystyle \frac {1}{4}(\pi +2t)$
0%
$\displaystyle \frac {1-\sin t}{\cos t}$
0%
$\displaystyle \frac {1}{4}(2t-\pi)$
0%
$\displaystyle \frac {1+\sin t}{\cos 2t}$

Explanation

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

$\dfrac{dx}{dt}=2acos^{2}t$

$y=a(1+sint)^2$

$\dfrac{dy}{dt}=2a(cost+costsint)$

$\Rightarrow \displaystyle \dfrac{dy}{dx}=\dfrac{1+sint}{cost}$

$\displaystyle = \dfrac{(cos\dfrac{t}{2}+sin\dfrac{t}{2})^{2}}{cos^{2}\dfrac{t}{2}-sin^{2}\dfrac{t}{2}}$

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

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

Slope of tangent to the curve

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

So, the angle made by the tangent to the curve with the x-axis

$=\dfrac{\pi}{4}+\dfrac{t}{2}$ or

$\dfrac{1}{4}(\pi+2t)$

An equation for the line that passes through

$(10, -1)$ and is perpendicular to

$y \displaystyle = \frac{x^2}{4} - 2$ is

0%
$4x + y = 39$
0%
$2x + y = 19$
0%
$x + y = 9$
0%
$x + 2y = 8$

Explanation

Given equation of parabola is

$y=\dfrac{x^2}{4}-2$

$\displaystyle\Rightarrow \frac{dy}{dx}=\frac{x}{2}$
Slope of tangent to parabola at

$\displaystyle(x_1,y_1)=\frac{x_1}{2}$
Therefore, slope of normal at

$(x_1,y_1)= \displaystyle - \frac{2}{x_1}$
Also, slope of normal

$= \displaystyle \frac{y_1 + 1}{x_1 - 10}$

$\therefore \displaystyle \frac{y_1 + 1}{x_1 - 10} = - \frac{2}{x_1}$

$\Rightarrow x_1 y_1 + x_1 = -2x_1 + 20$

$\Rightarrow x_1 y_1 + 3x_1 = 20$
Substituting

$y_1 \displaystyle = \frac{x_1^2 - 8}{4}$ (from the given equation)

$x_1 \displaystyle \left ( \frac{x_1^2 - 8}{4} + 3 \right ) = 20$

$\Rightarrow x_1 (x_1^2 + 4) = 80$

$\Rightarrow x_1^3 + 4x_1 - 80 = 0$ .
which has one root

$x_1 = 4$
Hence,

$x_1 = 4; y_1 = 2$

$\therefore P = (4, 2)$
Therefore, equation of

$PA$ is

$y + 1 = - \dfrac{1}{2} (x - 10)$

$\Rightarrow 2y + 2 = -x + 10$

$\Rightarrow x + 2y - 8 = 0$

The abscissas of points

$P$ and

$Q$ on the curve

$y=e^x+e^{-x}$ such that tangents at

$P$ and

$Q$ make

$60^{\circ}$ with the

$x$ -axis are

0%
$\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {7}\right )$ and $\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 5} {2}\right )$
0%
$\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {2}\right )$
0%
$\ln \left (\displaystyle \frac {\sqrt 7-\sqrt 3} {7}\right )$
0%
$\pm\ln \left (\displaystyle \frac {\sqrt 3+\sqrt 7} {2}\right )$

Explanation

$y=e^{x}+e^{-x}$

$\displaystyle\frac{dy}{dx}=e^{x}-e^{-x}=\frac{e^{2x}-1}{e^{x}}$
Let P

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

$(x_2,y_2)$ be the points on the given curve.
Slope of tangent at P is

$\displaystyle m_1=\frac{e^{2x_{1}}-1}{e^{x_{1}}}$

$\Rightarrow \displaystyle \sqrt{3}=\frac{e^{2x_{1}}-1}{e^{x_{1}}}$

$\displaystyle\Rightarrow e^{2x_1}-\sqrt{3}e^{x}-1=0$

$\displaystyle\Rightarrow (e^{x_1}-\frac{\sqrt{3}}{2})^{2}=\frac{7}{4}$

$\displaystyle\Rightarrow e^{x_1}=\pm\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2}$

$\displaystyle\Rightarrow {x_1}=ln{(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2})}$ as logarithm of negative numbers is not defined.
Similarly,

$\displaystyle\Rightarrow {x_2}=ln{(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2})}$ as it makes same angle,

$60^{0}$ with x-axis.

If the curve represented parametrically by the equations

$x=2 \ln\cot t+1$ and

$y=\tan t+ \cot t$

0%
tangent and normal intersect at the point $(2,1)$
0%
normal at $t=\displaystyle \frac{\pi}{4}$ is parallel to the $y$ axis
0%
tangent at $t=\displaystyle \frac{\pi}{4}$ is parallel to the line $y=x$
0%
tangent at $t=\displaystyle \frac{\pi}{4}$ is parallel to the $x$ axis

Explanation

$x=2 \ln\cot t+1$

$\displaystyle \frac{dx}{dt}=-\frac{2cosec^{2}t}{\cot t}=-\frac{2}{\sin t\cos t}$

$y=\tan t+ \cot t$

$\displaystyle \frac{dy}{dt}=sec^{2}t-cosec^{2}t=\frac{sin^{2}t-cos^{2}t}{sin^{2}t cos^{2}t}$

$\Rightarrow \displaystyle \frac{dy}{dx}=-\frac{sin^{2}t-cos^{2}t}{sin2t}$

$\Rightarrow \displaystyle \frac{dy}{dx}=\cot 2t$
Slope of tangent at

$t=\dfrac{\pi}{4}$ is 0.
Hence, tangent is parallel to x-axis.

A curve is represented by the equations,

$\displaystyle x = \sec^2t$ and

$y =\cot t$ , where

$t$ is a parameter. If the tangent at the point

$P$ on the curve where

$\displaystyle t = \dfrac{\pi}{4}$ meets the curve again at the point

$Q$ then

$|PQ|$ is equal to

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$\displaystyle \frac {5\sqrt 3}{2}$
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$\displaystyle \frac {5\sqrt 5}{2}$
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$\displaystyle \frac {2\sqrt 5}{3}$
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$\displaystyle \frac {3\sqrt 5}{2}$

Explanation

$x=\sec^{2}t$

$y=\cot t$

$\dfrac{dy}{dx}=-\dfrac{\csc^{2}t}{2\sec^{2}t\cdot\tan t}=-\dfrac{\cot^{3}t}{2}$

Now,

$\left.\dfrac{dy}{dx}\right|_{t=45^{0}}=-\dfrac{1}{2}$

$\left.x\right|_{t=45^{0}}=2$

$\left.y\right|_{t=45^{0}}=1$

Hence the equation of the tangent will be

$y-1=-\dfrac{1}{2}(x-2)\Rightarrow2y-2=-x+2\Rightarrow x+2y=4$ ...(i)

Now,

$\sec^{2}t=1+\tan^{2}t$

$\sec^{2}t=1+\dfrac{1}{\cot^{2}t}$
Or

$x=1+\dfrac{1}{y^{2}}\Rightarrow xy^{2}=y^{2}+1$ is the equation of curve.

Solving the equation of tangent and equation of curve we get

$4-2y=1+\dfrac{1}{y^{2}}$
Or

$4y^{2}-2y^{3}=y^{2}+1\Rightarrow2y^{3}-3y^{2}+1=0$

$y=-\dfrac{1}{2}$ and

$y=1$

For

$y=1; x= 2$

For

$y=-\dfrac{1}{2}; x = 5$

Hence,

$Q=\left(5,-\dfrac{1}{2}\right)$ and

$P=(2,1)$

$\therefore PQ=\dfrac{3\sqrt{5}}{2}$ .

The real number

$\displaystyle '\alpha'$ such that the curve

$\displaystyle f(x) = e^x$ is tangent to the curve

$\displaystyle g(x) = \alpha x^2$ .

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$\displaystyle \frac{e^2}{4}$
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$\displaystyle \frac{e^2}{2}$
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$\displaystyle \frac{e}{4}$
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$\displaystyle \frac{e}{2}$

Explanation

At the point of contact both the curves must have the same slope.
Hence

$f'(x)_{x=x_{1}}=g'(x)_{x=x_{1}}$
Or

$e^{x_{1}}=2\alpha.x_{1}$
Or

$\alpha=\dfrac{e^{x_{1}}}{2x_{1}}$
Now, considering

$x_{1}=1$ and

$x_{1}=2$ we get

$\alpha=\dfrac{e}{2}$ and

$\alpha=\dfrac{e^{2}}{4}$ .

The x-intercept of the tangent at any arbitrary point of the curve

$\displaystyle \frac {a}{x^2} + \frac {b}{y^2} = 1$ is proportional to:

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square of the abscissa of the point of tangency
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square root of the abscissa of the point of tangency
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cube of the abscissa of the point of tangency
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cube root of the abscissa of the point of tangency

The angle made by the tangent of the curve

$\displaystyle x = a(t + \sin t \cos t); y = a (1 + \sin t)^2$ with the x-axis at any point on it is

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$\displaystyle \frac {1}{4} (\pi + 2t)$
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$\displaystyle \frac {1 - \sin t}{ \cos t}$
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$\displaystyle \frac {1}{4} (2t - \pi)$
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$\displaystyle \frac {1 + \sin t}{\cos 2 t}$

Explanation

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

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

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

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

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

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

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

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

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

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

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

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

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

$=\dfrac{\left(\cos{\dfrac{t}{2}}+\sin{\dfrac{t}{2}}\right)}{\left(\cos{\dfrac{t}{2}}-\sin{\dfrac{t}{2}}\right)}\\$

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

$=\dfrac{\tan{\dfrac{\pi}{4}}+\tan{\dfrac{t}{2}}}{1-\tan{\dfrac{\pi}{4}}\tan{\dfrac{t}{2}}}\\$

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

$\theta$ is the angle which the tangent at any point

$t$ makes with the

$X$ axis

then

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

$\therefore\,\theta=\dfrac{\pi}{4}+\dfrac{t}{2}=\dfrac{1}{4}\left(\pi+2t\right)$

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

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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