Tangents and its Equations Test 10

Result

Tangents and its Equations Test 10

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

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

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

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

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

$$d=\dfrac{2}{\sqrt{5}}$$
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 $$\cos x<0$$
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$$^{n}$$
$$\sin^{4}\ \theta=\cos^{4}\ \theta$$
$$\theta=\dfrac {\pi}4$$
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

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

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 10

Result

Tangents and its Equations Test 10

Score

-

Rank

-

SHARING IS CARING

Question 1

$$x+y=e$$

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

$$y+ ex=1$$

$$x+ey=2$$

Solution

Question 2

$$2x-y=3a$$

$$2x+y=2a$$

$$x+2y=6a$$

$$x+y=a$$

Solution

Question 3

$$ax+ by=1$$

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

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

$$ax- by=1$$

Solution

Question 4

Trisected at that point

bisected at that point

constant

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

Solution

Question 5

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

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

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

$$1$$

Solution

Question 6

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

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

$$2\sqrt{5}$$

$$5\sqrt{2}$$

Solution

Question 7

$$C, B, D, A$$

$$B, C, A, D$$

$$A, D, C, B$$

$$D, A, C, B$$

Solution

Question 8

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

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

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

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

Solution

Question 9

$$y=\pm x$$

$$y=\pm 2x$$

$$y=\pm 3x$$

$$x=\pm 2y$$

Solution

Question 10

$$x=y$$

$$x=0$$

$$y=0$$

$$x+y=0$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.