Tangents and its Equations Test 20

Result

Tangents and its Equations Test 20

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

A and B are points $$(-2,0)$$ and $$(1,3)$$ on the curve $$\displaystyle y=4-x^{2}$$. If the tangent at P on the curve be parallel to chord AB, then co-ordinates of point P are

A
$$\displaystyle \left ( -\frac{1}{3}, \frac{5}{3} \right )$$

B
$$\displaystyle \left ( \frac{1}{2}, -\frac{15}{4} \right )$$

C
$$\displaystyle \left ( -\frac{1}{2}, \frac{15}{4} \right )$$

D
$$\displaystyle \left ( -\frac{1}{3}, \frac{1}{5} \right )$$

Solution

Given equation of curve $$\displaystyle y=4-x^{2}$$ ...(i)
$$\dfrac{dy}{dx}=-2x$$

Given points $$A(-2,0)$$ and $$(1,3)$$
Slope of AB $$=\dfrac{3}{3}=1$$

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

So, by equation (i), we get
$$y=\dfrac{15}{4}$$
Hence, the point is $$(-\dfrac{1}{2},\dfrac{15}{4})$$
Question 2
1 / -0

The parametric equations of a curve are given by $$\displaystyle x= \sec ^{2}t, y= \cot t.$$ Tangent at $$\displaystyle P \,t=\dfrac { \pi }4$$ meets the curve again at Q; then $$\displaystyle PQ=? $$

A
$$ \left ( 2\sqrt{5} \right ).$$

B
$$\dfrac {\left ( 5\sqrt{5} \right )}2.$$

C
$$ \dfrac {\left ( 3\sqrt{5} \right )}2.$$

D
$$ \left ( 3\sqrt{5} \right ).$$

Solution

$$\displaystyle x= \sec ^{2}t, y= \cot t.$$
Given $$t=\dfrac{\pi}{4}$$
$$\Rightarrow x=\sec^2 (\dfrac{\pi}{4}), y=\cot \dfrac{\pi}{4}$$
$$\Rightarrow x=2, y=1$$
So, the point P is $$(2,1)$$

Now, $$\dfrac{dx}{dt}=2\sec t \sec t\tan t$$
$$\dfrac{dy}{dt}=-cosec^2 t$$
$$\Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{2}\cot^3 t$$
Slope of tangent at P is $$-\dfrac{1}{2}$$

Equation of tangent at P is
$$y-1=-\dfrac{1}{2}(x-2)$$
$$\Rightarrow x+2y=4$$ .....(1)

Since, $$\sec^2 t -\tan^2 t=1$$
$$\Rightarrow x-\dfrac{1}{y^2}=1$$ .....(2)

Solving (1) and (2), we get
$$\displaystyle \left ( 4-2y \right )y^{2}-1= y^{2}.$$
$$\Rightarrow \displaystyle y= 1, 1,-\frac{1}{2}$$
$$\Rightarrow x=2, 5$$ (by (1))

$$\displaystyle \therefore Q is\left ( 5,-\frac{1}{2} \right ) $$
$$\therefore PQ= \sqrt{\dfrac{45}{4}}= \dfrac{3\sqrt{5}}{2}.$$
The values of $$y=1,1$$ correspond to point P.
Question 3
1 / -0

The line $$\dfrac xa+\dfrac yb=1$$ touches the curve $$\displaystyle y=be^{-x/a}$$ at the point

A
$$(a,b/a)$$

B
$$(-a,b/a)$$

C
$$(a,a/b)$$

D
None of these

Solution

Simplifying the equation of the line we get
$$bx+ay=ab$$
$$ay=-bx+ab$$
Or
$$y=\dfrac{-b}{a}.x+b$$
Hence
$$\dfrac{dy}{dx}=\dfrac{-b}{a}$$
Or
$$-\dfrac{b}{a}.e^{-x/a}.=\dfrac{-b}{a}$$
Or
$$e^{-x/a}=1$$
Or
$$x=0$$
Hence
$$y=b$$.
Therefore the point is $$(0,b)$$
Question 4
1 / -0

The curve $$\displaystyle y-e^{xy}+x=0$$ has a vertical tangent at the point

A
$$(1,\ 1)$$

B
$$no\ point$$

C
$$(0,\ 1)$$

D
$$(1,\ 0)$$

Solution

$$\dfrac{dy}{dx}-e^{xy}[x\dfrac{dy}{dx}+y]+1=0$$
Or
$$\dfrac{dy}{dx}[1-x.e^{xy}]+1-y.e^{xy}=0$$
Or
$$\dfrac{dy}{dx}=\dfrac{-(1-y.e^{xy})}{1-x.e^{xy}}$$
Now
$$1-xe^{xy}=0$$ since the slope of the tangent is $$90^{0}$$.
Hence
$$xe^{xy}=1$$
Or
$$x(x+y)=1$$
Or
$$x^{2}+xy=1$$
Or
$$y=\dfrac{1-x^{2}}{x}$$
Or
$$y=\dfrac{1}{x}-x$$ ...(i)
Considering $$y=0$$,
$$x^{2}=1$$
$$x=\pm1$$.
Hence we get 2 points $$(1,0)$$ and $$(-1,0)$$.
Out of these 2, only $$(1,0)$$ lies on the curve.
Hence the required point is $$(1,0)$$.
Question 5
1 / -0

If $$\displaystyle x\cos \alpha +y\sin \alpha =p$$ touches $$\displaystyle x^{2}+a^{2}y^{2}=a^{2},$$ then

A
$$\displaystyle p^{2}=a^{2}\sin^{2}\alpha +\cos^{2}\alpha $$

B
$$\displaystyle p^{2}=a^{2}\cos^{2}\alpha +\sin^{2}\alpha $$

C
$$\displaystyle 1/p^{2}=\sin^{2}\alpha +\alpha^{2}\cos^{2}\alpha $$

D
$$\displaystyle 1/p^{2}=\cos^{2}\alpha +\alpha^{2}\sin^{2}\alpha $$

Solution

Solving for y, we have
$$\displaystyle y=\dfrac { p }{ \sin { \alpha } } -x\cot { \alpha } $$
Putting this value in $${ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }={ a }^{ 2 }$$, we have
$$\displaystyle { x }^{ 2 }+{ a }^{ 2 }{ \left( \dfrac { p }{ \sin { \alpha } } -x\cot { \alpha } \right) }^{ 2 }={ a }^{ 2 }$$

$$\displaystyle \Rightarrow { x }^{ 2 }\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) -\dfrac { 2a^{ 2 }xp\cot { \alpha } }{ \sin { \alpha } } +\dfrac { { a }^{ 2 }{ p }^{ 2 } }{ \sin ^{ 2 }{ \alpha } } -{ a }^{ 2 }=0$$

The discriminant of this equation must be zero. So
$$\displaystyle { a }^{ 4 }\dfrac { { p }^{ 2 }\cot ^{ 2 }{ \alpha } }{ \sin ^{ 2 }{ \alpha } } =\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) \left( \dfrac { { a }^{ 2 }{ p }^{ 2 } }{ \sin ^{ 2 }{ \alpha } } -{ a }^{ 2 } \right) \\ \Rightarrow { a }^{ 2 }{ p }^{ 2 }\cot ^{ 2 }{ \alpha } =\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) \left( { p }^{ 2 }-\sin ^{ 2 }{ \alpha } \right) \\ \Rightarrow { p }^{ 2 }\left( { a }^{ 2 }\cot ^{ 2 }{ \alpha } -1-{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) =\sin ^{ 2 }{ \alpha } -{ a }^{ 2 }\cos ^{ 2 }{ \alpha } \\ \Rightarrow { p }^{ 2 }={ a }^{ 2 }\cos ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \alpha } $$
Question 6
1 / -0

The point of intersection of the tangents drawn to the curve $$\displaystyle x^{2}y=1-y$$ at the points where it is met by the curve $$\displaystyle xy=1-y$$ is given by

A
$$\displaystyle \left ( 0, -1\right )$$

B
$$\displaystyle \left ( 1, 1\right )$$

C
$$\displaystyle \left ( 0, 1\right )$$

D
none of these

Solution

Intersection of $${ x }^{ 2 }y=1-y$$ are $$xy=1-y$$
are given by $${ x }^{ 2 }y=xy$$.
Since $$y\neq 0$$ on any of two curves
So we must have $$x=0,1$$.
Thus the curve intersect at $$\displaystyle \left( 0,1 \right) ;\left( 0,\dfrac { 1 }{ 2 } \right) $$

Now differentiating $${ x }^{ 2 }y=1-y$$. we have
$$\displaystyle \dfrac { dy }{ dx } =\dfrac { 2xy }{ 1+{ x }^{ 2 } } $$
Thus for $$\left( x,y \right) =\left( 0,1 \right) $$ $$\displaystyle \dfrac { dy }{ dx } =0$$
And for $$\displaystyle \left( x,y \right) =\left( 0,\dfrac { 1 }{ 2 } \right) $$$$\displaystyle \dfrac { dy }{ dx } =-\dfrac { 1 }{ 2 } $$
Thus the equation of tangent are $$Y=1$$ and $$\displaystyle Y-\dfrac { 1 }{ 2 } =\left( -\dfrac { 1 }{ 2 } \right) \left( X-1 \right) $$
Their intersection is given by $$\left( 0,1 \right) $$
Question 7
1 / -0

The coordinates of the point $$M(x, y)$$ on $$\displaystyle y= e^{-\left | x \right |}$$ so that the area formed by the coordinates axes and the tangent at $$M$$ is greatest, are

A
$$(e, 1)$$

B
$$\displaystyle \left (1, e^{-1} \right )$$

C
$$\displaystyle \left (-1, e^{-1} \right )$$

D
$$(0, 1)$$

Solution

The curve of f(x) is symmetrical about y axis.
Let the point of contact of the tangent be $$(a,e^{-a})$$
Therefore
$$\dfrac{dy}{dx}_{(a,e^{-a})}=-e^{-a}$$
Hence the equation of the tangent will be
$$y-e^{-a}=-e^{-a}(x-a)$$

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

$$y+e^{-a}x=e^{-a}(1+a)$$

$$\dfrac{y}{e^{-a}(1+a)}+\dfrac{x}{1+a}=1$$
Hence the area formed by the tangent and the coordinate axes will be
$$=\dfrac{(x-intercept)(y-intercept)}{2}$$

$$A=\dfrac{e^{-a}(1+a)^{2}}{2}$$

Differentiating with respect to a, we get
$$\dfrac{dA}{da}=\dfrac{1}{2}[2(1+a)e^{-a}-(1+a)^{2}e^{-a}]$$

$$=\dfrac{(1+a)e^{-a}}{2}[2-(1+a)]$$

$$=\dfrac{(1+a)e^{-a}}{2}(1-a)$$
$$=0$$
Hence
$$a=\pm1$$
Thus we get the points as $$(1,e^{-1})$$ and $$(-1,e^{-1})$$.
Question 8
1 / -0

The value of m for which the area of the triangle included between the axes and any tangent to the $$\displaystyle x^{m}y= b^{m}$$ curve is constant, is

A
$$\dfrac 12$$

B
$$1$$

C
$$\dfrac 32$$

D
$$2$$

Solution

Given, $$y =b^mx^{-m}$$
$$y' = -b^mx^{-(m+1)}$$
At any point $$(x1,y1)$$
Equation of tangent is given by
$$\cfrac{y-y1}{x-x1} = -b^mx1^{-(m+1)}$$
Y intercept $$= -m$$
X intercept $$= (m-1)\times \dfrac 1m$$
Area bounded $$= \cfrac{1}{2}\cfrac{(m-1)x1}{m} \times m$$
For area to be constant, $$m =1$$
Question 9
1 / -0

If the tangent at any point on the curve $$\displaystyle x^{4}+y^{4}= a^{4}$$ cuts off intercepts p and q on the coordinate axes, the value of $$\displaystyle p^{-\tfrac 43}+q^{-\tfrac 43}$$ is

A
$$\displaystyle a^{-\tfrac 43}$$

B
$$\displaystyle a^{-\tfrac 12}$$

C
$$\displaystyle a^{\tfrac 12}$$

D
none of these

Solution

Let
$$x=a\sqrt{\cos \theta}$$
$$y=a\sqrt{\sin \theta}$$
Hence
$$\dfrac{dy}{dx}$$

$$=-\dfrac{\cos \theta.\sqrt{\cos \theta}}{\sin \theta.\sqrt{\sin \theta}}$$
$$=\cot\theta^{\dfrac{3}{2}}$$.
Hence equation of tangent will be
$$y-a\sqrt{\sin \theta}=-\dfrac{\cos \theta.\sqrt{\cos \theta}}{\sin \theta.\sqrt{\sin \theta}}(x-a\sqrt{\cos \theta})$$
$$\sin \theta.\sqrt{\sin \theta}y-a\sin ^{2}\theta=-\cos \theta\sqrt{\cos \theta}x+a\cos ^{2}\theta$$
$$(\sin \theta)^{\dfrac{3}{2}}.y+(\cos \theta)^{\dfrac{3}{2}}x=a$$
Hence the intercepts are
$$Q=\dfrac{a}{(\sin \theta)^{\dfrac{3}{2}}}$$ and $$P=\dfrac{a}{(\cos \theta)^{\dfrac{3}{2}}}$$
Hence
$$Q^{\dfrac{-4}{3}}+P^{\dfrac{-4}{3}}$$
$$=a^{-\dfrac{4}{3}}[\cos ^{2}\theta+\sin ^{2}\theta]$$
$$=a^{-\dfrac{4}{3}}$$.
Question 10
1 / -0

The equation of the tangent to the curve $$\displaystyle y= \left ( 2x-1 \right )e^{2\left ( 1-x \right )}$$ at the point of its maximum is

A
$$y=1$$

B
$$x=1$$

C
$$x+y=1$$

D
$$x-y=-1$$

Solution

Given, $$y=(2x-1)e ^{2(1-x)}$$
$$\therefore y'\left( x \right) =-2\left( 2x-1 \right) { e }^{ 2\left( 1-x \right) }+2{ e }^{ 2\left( 1-x \right) }\\ =2{ e }^{ 2\left( 1-x \right) }\left( -2x+2 \right) $$
Thus $$y'\left( x \right) =0\Rightarrow x=1$$
Since $$y''\left( 1 \right) <0$$ so $$\left( 1,1 \right) $$ is the point of maximum and the equation of tangent is
$$y-1=0\left( x-1 \right)$$ ,i.e. $$y=1$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 20

Result

Tangents and its Equations Test 20

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\displaystyle \left ( -\frac{1}{3}, \frac{5}{3} \right )$$

$$\displaystyle \left ( \frac{1}{2}, -\frac{15}{4} \right )$$

$$\displaystyle \left ( -\frac{1}{2}, \frac{15}{4} \right )$$

$$\displaystyle \left ( -\frac{1}{3}, \frac{1}{5} \right )$$

Solution

Question 2

$$ \left ( 2\sqrt{5} \right ).$$

$$\dfrac {\left ( 5\sqrt{5} \right )}2.$$

$$ \dfrac {\left ( 3\sqrt{5} \right )}2.$$

$$ \left ( 3\sqrt{5} \right ).$$

Solution

Question 3

$$(a,b/a)$$

$$(-a,b/a)$$

$$(a,a/b)$$

None of these

Solution

Question 4

$$(1,\ 1)$$

$$no\ point$$

$$(0,\ 1)$$

$$(1,\ 0)$$

Solution

Question 5

$$\displaystyle p^{2}=a^{2}\sin^{2}\alpha +\cos^{2}\alpha $$

$$\displaystyle p^{2}=a^{2}\cos^{2}\alpha +\sin^{2}\alpha $$

$$\displaystyle 1/p^{2}=\sin^{2}\alpha +\alpha^{2}\cos^{2}\alpha $$

$$\displaystyle 1/p^{2}=\cos^{2}\alpha +\alpha^{2}\sin^{2}\alpha $$

Solution

Question 6

$$\displaystyle \left ( 0, -1\right )$$

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

$$\displaystyle \left ( 0, 1\right )$$

none of these

Solution

Question 7

$$(e, 1)$$

$$\displaystyle \left (1, e^{-1} \right )$$

$$\displaystyle \left (-1, e^{-1} \right )$$

$$(0, 1)$$

Solution

Question 8

$$\dfrac 12$$

$$1$$

$$\dfrac 32$$

$$2$$

Solution

Question 9

$$\displaystyle a^{-\tfrac 43}$$

$$\displaystyle a^{-\tfrac 12}$$

$$\displaystyle a^{\tfrac 12}$$

none of these

Solution

Question 10

$$y=1$$

$$x=1$$

$$x+y=1$$

$$x-y=-1$$

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.