Tangents and its Equations Test 10

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Tangents and its Equations Test 10

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Weekly Quiz Competition

Question 1
1 / -0

The equation of the tangent to the curve $y=e^{-|x|}$ at the point where the curve cuts the line $x=1$ is

A
$x+y=e$

B
$e(x+y)=1$

C
$y+ ex=1$

D
$x+ey=2$

Solution

$y=e^{-x}$
$y'=\dfrac{-1e^{-x}}{e^{-x}}$
$y'=-e^{-1} \ \ at \ \ x=1$
$y=e^{-1} \ \ at \ \ x=1$
$(y-e^{-1})=-e^{-1}(x-1)$
$x+ey=2$

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Question 2
1 / -0

The equation of the normal at $x$ $=$ $2a$ for the curve $\displaystyle y=\frac{8a^{3}}{4a^{2}+x^{2}}$ is

A
$2x-y=3a$

B
$2x+y=2a$

C
$x+2y=6a$

D
$x+y=a$

Solution

Equation of normal at $x=2a$ for $y=\dfrac { 8{ a }^{ 3 } }{ 4{ a }^{ 2 }+{ x }^{ 2 } }$
$\Rightarrow x=2a$
$\Rightarrow y=\dfrac { 8{ a }^{ 3 } }{ 4{ a }^{ 2 }+{ (2a) }^{ 2 } } =a$
$y'=\dfrac { dy }{ dx } =\dfrac { -8{ a }^{ 3 }(2x) }{ { \left( 4{ a }^{ 2 }+{ x }^{ 2 } \right) }^{ 2 } } ,x=2a\Rightarrow y'=\dfrac { -8{ a }^{ 3 }\times 2\times 2a }{ { \left( 4{ a }^{ 2 }+4{ a }^{ 2 } \right) }^{ 2 } }$
$=\dfrac { -8{ a }^{ 3 }\times 4a }{ 64{ a }^{ 4 } }$
$=-\dfrac { 1 }{ 2 }$
Slope of normal $=\dfrac { -1 }{ y' } =2$
$\Rightarrow y-a=2(x-2a)$
$2x-4a=y-a$
$\therefore 2x-y=3a$ is a equation of normal.

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Question 3
1 / -0

Equation of the tangent line to $y=be^{\frac{-x}{a}}$ where it crosses y-axis is

A
$ax+ by=1$

B
$\displaystyle \frac{x}{a}+\frac{y}{b}=1$

C
$\displaystyle \frac{x}{b}+\frac{y}{a}=1$

D
$ax- by=1$

Solution

$y=b e^{\frac{-x}{a}} at x=0$
$y=b$
$y'=\dfrac{-b e^{-x}}{a}$
$y'=\dfrac{-b }{a} \ \ (at \ \ x=0)$
$(y-b )=\dfrac{-b }{a}(x-0)$
$\dfrac{x}{a}+\dfrac{y}{b }=1$

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Question 4
1 / -0

The portion of the tangent to xy $=a^{2}$ at any point on it between the axes is

A
Trisected at that point

B
bisected at that point

C
constant

D
with ratio $1 : 4$ at the point

Solution

Let xy tangent at $(a,a)$

$xy'+y=0$

$y^{1}=\cfrac{-y}{x}$

$y'=-1$

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

Bisected by that point.

$y' \ \ at \ \ B =\dfrac{-y}{x}$

$=\dfrac{-ay^{2}}{ty^{2}}$

$(y-\dfrac{a^{2}}{t})=\dfrac{-a^{2}}{t^{2}}(x-t)$

$x=0 , y=\dfrac{2a^{2}}{t}$

$y=0, x=2t$

$(t,\dfrac{a^{2}}{t})$ to the mid point of (2+10) and $(0,\dfrac{2a^{2}}{t})$ .
H.P.

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Question 5
1 / -0

Area of the triangle formed by the normal to the curve $x=e^{\sin y}$ at (1, 0) with the coordinate axes is

A
$\displaystyle \frac{1}{4}$

B
$\displaystyle \frac{1}{2}$

C
$\displaystyle \frac{3}{4}$

D
$1$

Solution

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Question 6
1 / -0

The distance of the origin from the normal to the curve $y=e^{2x}+x^{2}$ at $x=0$ is

A
$\displaystyle \frac{2}{5}$

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

C
$2\sqrt{5}$

D
$5\sqrt{2}$

Solution

$y'=2e^{2x}+2x$
$y=1$ at $x=0$
$y'=2$

$m$ of normal $=\dfrac{-1}{2}$

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

$x+2y-2=0$

distance from origin

$d=\left |\dfrac{x_{1}+2y_{1}-2}{\sqrt{1^{2}+2^{2}}} \right | (x_{1}y_{1})=(0,0)$

$d=\dfrac{2}{\sqrt{5}}$

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Question 7
1 / -0

The arrangment of the slopes of the normals to the curve $y=e^{\log(cosx)}$ in the ascending order at the points given below.
$A) \displaystyle x=\frac{\pi}{6}, B) \displaystyle x=\frac{7\pi}{4}, C)x=\frac{11\pi}{6}, D)x=\frac{\pi}{3}$

A
$C, B, D, A$

B
$B, C, A, D$

C
$A, D, C, B$

D
$D, A, C, B$

Solution

We have, $y = e^{\log \cos x} =(\cos x)^{\log e}[\because a^{\log_bc} =c^{\log_ba}] =\cos x$
$\Rightarrow \dfrac{dy}{dx} =-\sin x=m$ (say)
Thus slope of normal to the given curve at any point is,
$m'=- \dfrac{1}{m}=\dfrac{1}{\sin x}$
Now, $m'_{\frac{\pi}{6}}=2, m'_{\frac{7\pi}{4}}=-\sqrt{2}, m'_{\frac{\pi}{6}}=-2,m'_{\frac{\pi}{6}}=\dfrac{\sqrt{3}}{2}$
Clearly ascending order is, $C,B,D,A$
Note: Given function is not defined where $\cos x<0$

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Question 8
1 / -0

lf the normal at the point $p(\theta)$ of the curve $x^{\tfrac{2}{3}}+y^{\tfrac{2}{3}}=a^{\tfrac{2}{3}}$ passes through the origin then

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

B
$\theta =\dfrac {\pi}6$

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

D
$\theta =\dfrac {\pi}2$

Solution

$x=a\ \cos^{3}\theta$
$y=a\ \sin^{3}\theta$
$y^{1}=\dfrac{a\ 3\ \sin^{2}\theta\ \cos\ \theta}{a\ 3\ \cos^{2}\theta-\cos\ \theta}$
$\dfrac{-\sin\ \theta}{\cos\ \theta}(y-a\ \sin^{3}\theta)=\dfrac{\cos\ \theta}{\sin\ \theta}(x-a\ \cos^{3}\ \theta)$
$(0, 0)$ satisfies the eq $^{n}$
$\sin^{4}\ \theta=\cos^{4}\ \theta$
$\theta=\dfrac {\pi}4$

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Question 9
1 / -0

The equations of the tangents at the origin to the curve $y^{2}=x^{2}(1+x)$ are

A
$y=\pm x$

B
$y=\pm 2x$

C
$y=\pm 3x$

D
$x=\pm 2y$

Solution

$y=+_-\sqrt{x^{2}+x^{3}}$
$y'=+_-\dfrac{2x+3x^{2}}{2\sqrt{x^{2}+x^{3}}}$
$y'=+_-\dfrac{2+3x}{2\sqrt{1+x}}$
$y'|_{x=0}=+_-1$
$y=+_-x$

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Question 10
1 / -0

The equation of the common normal at the point of contact of the curves $x^{2}=y$ and $x^{2}+y^{2}-8y=0$

A
$x=y$

B
$x=0$

C
$y=0$

D
$x+y=0$

Solution

$y^{2}+y-8y=0$
$y=0.\ y=7$
$x=0,\ x=\sqrt{7}$
$y^{1}=2x=m_{1}$
$yy^{1}-4y^{1}=-x$
$y^{1}=\frac{x}{4-y}=m_{2}$
$m_{1},\ m_{2}\ at\ (0,0)$
$m_{1}=m_{2}=0$
Slope of the normal $=\alpha$
$\therefore x=0$ is the required line
cl. y-axis

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Tangents and its Equations Test 10

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Question 1

x+y=ex+y=ex+y=e

e(x+y)=1e(x+y)=1e(x+y)=1

y+ex=1y+ ex=1y+ex=1

x+ey=2x+ey=2x+ey=2

Solution

Question 2

2x−y=3a2x-y=3a2x−y=3a

2x+y=2a2x+y=2a2x+y=2a

x+2y=6ax+2y=6ax+2y=6a

x+y=ax+y=ax+y=a

Solution

Question 3

ax+by=1ax+ by=1ax+by=1

xa+yb=1\displaystyle \frac{x}{a}+\frac{y}{b}=1ax​+by​=1

xb+ya=1\displaystyle \frac{x}{b}+\frac{y}{a}=1bx​+ay​=1

ax−by=1ax- by=1ax−by=1

Solution

Question 4

Trisected at that point

bisected at that point

constant

with ratio 1:41 : 41:4 at the point

Solution

Question 5

14\displaystyle \frac{1}{4}41​

12\displaystyle \frac{1}{2}21​

34\displaystyle \frac{3}{4}43​

111

Solution

Question 6

25\displaystyle \frac{2}{5}52​

25\displaystyle \frac{2}{\sqrt{5}}5​2​

252\sqrt{5}25​

525\sqrt{2}52​

Solution

Question 7

C,B,D,AC, B, D, AC,B,D,A

B,C,A,DB, C, A, DB,C,A,D

A,D,C,BA, D, C, BA,D,C,B

D,A,C,BD, A, C, BD,A,C,B

Solution

Question 8

θ=π3\theta =\dfrac {\pi}3θ=3π​

θ=π6\theta =\dfrac {\pi}6θ=6π​

θ=π4\theta =\dfrac {\pi}4θ=4π​

θ=π2\theta =\dfrac {\pi}2θ=2π​

Solution

Question 9

y=±xy=\pm xy=±x

y=±2xy=\pm 2xy=±2x

y=±3xy=\pm 3xy=±3x

x=±2yx=\pm 2yx=±2y

Solution

Question 10

x=yx=yx=y

x=0x=0x=0

y=0y=0y=0

x+y=0x+y=0x+y=0

Solution

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$x+y=e$

$e(x+y)=1$

$y+ ex=1$

$x+ey=2$

$2x-y=3a$

$2x+y=2a$

$x+2y=6a$

$x+y=a$

$ax+ by=1$

$\displaystyle \frac{x}{a}+\frac{y}{b}=1$

$\displaystyle \frac{x}{b}+\frac{y}{a}=1$

$ax- by=1$

with ratio $1 : 4$ at the point

$\displaystyle \frac{1}{4}$

$\displaystyle \frac{1}{2}$

$\displaystyle \frac{3}{4}$

$1$

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

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

$2\sqrt{5}$

$5\sqrt{2}$

$C, B, D, A$

$B, C, A, D$

$A, D, C, B$

$D, A, C, B$

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

$\theta =\dfrac {\pi}6$

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

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

$y=\pm x$

$y=\pm 2x$

$y=\pm 3x$

$x=\pm 2y$

$x=y$

$x=0$

$y=0$

$x+y=0$