Tangents and its Equations Test 8

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Tangents and its Equations Test 8

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Weekly Quiz Competition

Question 1
1 / -0

The equation of normal to the curve $$2y=3-{ x }^{ 2 }$$ at the point $$\left(1,1\right)$$

A
$$x+y+1=0$$

B
$$x+y=0$$

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

D
$$x-y=0$$

Solution

The curve is $$2y=3-{x}^{2}$$

Slope$$=2{y}^{\prime}=-2x$$ or $$\dfrac{dy}{dx}=-x$$

Slope at $$\left(1,1\right)$$ is $$\dfrac{dy}{dx}=-1$$

Equation of normal is $$y-1=\dfrac{-1}{-1}\left(x-1\right)$$

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

$$\Rightarrow x-y=0$$ is the equation of the normal.
Question 2
1 / -0

equation of tangent at $$(0,0)$$ for the equation $$y^2=16x$$

A
$$y=0 $$

B
$$x=0$$

C
$$x+y=0$$

D
$$x-y=0$$

Solution

The equation is $$y^2=16x$$
The point is $$(0,0)$$
The slope of tangent $$2y \dfrac{dy}{dx}=16\\\dfrac{dy}{dx}=\dfrac 8y\\\left.\dfrac{dy}{dx}\right|_{(0,0)}=\infty$$
Equation of tangent is $$y-0=\dfrac{8}{0}(x-0)\\x=0$$
Question 3
1 / -0

The intercept on x-axis made by tangent to the curve, $$\displaystyle y=\int _{ 0 }^{ x }{ \left| t \right| } dt,x\in R$$, which are parallel to the line $$y=2x$$, are equal to

A
$$\pm 1$$

B
$$\pm 2$$

C
$$\pm 3$$

D
$$\pm 4$$

Solution

$$y = \int_{0}^{x}{\left| t \right| dt}$$
$$\Rightarrow \cfrac{dy}{dx} = \left| x \right|$$
Since tangent to the curve is parallel to the line $$y = 2x$$.
$$\therefore \cfrac{dy}{dx} = 2$$
$$\Rightarrow x = \pm 2$$
Therefore,
$$y = \int_{0}^{\pm 2}{\left| t \right| dt}$$
$$\Rightarrow y = \pm 2$$
Therefore,
Equation of tangents are-
$$y - 2 = 2 \left( x - 2 \right)$$
$$y + 2 = 2 \left( x + 2 \right)$$
For $$x$$-intercept, substitute $$y = 0$$, we have
$$0 - 2 = 2 \left( x - 2 \right) \Rightarrow x = +1$$
$$0 + 2 = 2 \left( x + 2 \right) \Rightarrow x = -1$$
Hence the intercept on $$x$$-axis are $$\pm 1$$.
Question 4
1 / -0

The area of triangle formed by tangent and normal at point $$(\sqrt{3}, 1)$$ of the curve $$x^2+y^2=4$$ and x-axis is?

A
$$\dfrac{4}{\sqrt{3}}$$

B
$$\dfrac{2}{\sqrt{3}}$$

C
$$\dfrac{8}{\sqrt{3}}$$

D
$$\dfrac{5}{\sqrt{3}}$$

Solution

Slope of OP$$=\dfrac{1}{\sqrt{3}}$$, slope of PQ$$=-\sqrt{3}$$
$$y-1=-\sqrt{3}(x-\sqrt{3})=-\sqrt{3}x+3$$
$$\Rightarrow \sqrt{3}x+y=4$$ and $$Q\left(\dfrac{4}{\sqrt{3}}, 0\right)$$
$$\Delta OPQ=\dfrac{2}{\sqrt{3}}$$.
Question 5
1 / -0

A curve $$y=me^{mx}$$ where $$m > 0$$ intersects y-axis at a point $$P$$.
What is the equation of tangent to the curve at $$P$$ ?

A
$$y=mx+m$$

B
$$y=-mx+2m$$

C
$$y=m^2x+2m$$

D
$$y=m^2x+m$$

Solution

$$y = m {e}^{mx}, \; m > 0$$
Substituting $$x = 0$$, we have
$$y = m {e}^{m \cdot 0} = m$$
Therefore,
Slope $$= \cfrac{dy}{dx} = {m}^{2} {e}^{mx} = {m}^{2} {e}^{m \cdot 0} = {m}^{2}$$
Therefore, equation of tangent at $$P$$ is given by-
$$y - m = \cfrac{dy}{dx} \left( x - 0 \right)$$
$$\Rightarrow y - m = {m}^{2} x$$
$$\Rightarrow y = {m}^{2}x + m$$
Question 6
1 / -0

The tangent to the curve $$y=e^{kx}$$ at a point (0, 1) meets the x-axis at (q, 0) where $$a \epsilon \left [ -2,-1 \right ]$$ then $$k \epsilon $$

A
[-1/2,0]

B
[-1,-1/2]

C
[0,1]

D
[1/2, 1]

Solution
Question 7
1 / -0

If the slope of the tangent to the curve $$xy+ ax+ by=0$$ at the point $$(1, 1) $$ on it is $$2$$, then values of $$a$$ and $$b$$ are

A
$$1, 2$$

B
$$1, -2$$

C
$$-1, 2$$

D
$$-1, -2$$

Solution

$$a+b +1=0$$
$${xy}'+y+a+{b y}'=0$$
$${y}'=\dfrac{-(y+a)}{(x+b )}$$
$$-\dfrac{(a+1)}{(1+b )}=2$$
$$2+2b +a+1=0$$
$$2+2 b -b =0$$
$$b =-2$$
$$a=1$$
Question 8
1 / -0

If the slope of the tangent to the curve $$y = x^{3}$$ at a point on it is equal to the ordinate of the point then the point is

A
$$(27, 3)$$

B
$$(3, 27)$$

C
$$(3, 3)$$

D
$$(1, 1)$$

Solution

Differntiating the given equation
$$\dfrac{dy}{dx}=3x^{2}$$
$$3a^{2}=a^{3}$$
$$a=3$$
$$(3,3^{3})$$
$$(3,27)$$
Question 9
1 / -0

The area of the triangle formed by the tangent to the curve $$\displaystyle y=\frac{8}{4+x^{2}}$$ at $$x=2$$ on it and the $$x$$-axis is

A
$$2$$ sq.units

B
$$4$$ sq.units

C
$$8$$ sq.units

D
$$16$$ sq.units

Solution

$${y}'=\dfrac{8\times 24}{(4+x^{2})^{2}}$$
$$y=1=\dfrac{-8\times 2\times 2 }{(4+4)^{2}}$$
$$=\dfrac{-5\times 4}{64}=\dfrac{1}{2}$$
$$(y-1)=\dfrac{-1}{2}(x-2)$$
$$2y-2=-x+2$$
$$2y+x=4$$
$$A=\dfrac{1}{2}\times 2\times 4=4$$
Question 10
1 / -0

The two curves $$y=x^{2}-1$$ and $$y=8x-x^{2}-9$$ at the point $$(2, 3)$$ have common

A
tangent as $$4x-y-5=0$$

B
tangent as $$x+4y-14=0$$

C
normal as $$4x+y=11$$

D
normal as $$x-4y=10$$

Solution

$$y_1 = x^2-1$$
$$\therefore y'_1=2x$$
$$y_2 = 8x-x^2-9$$
$$\therefore y'_2=8-2x$$

At the common tangent point, slope will be same.
$$2x=8-2x$$
$$4x=8$$
$$x=2, \therefore y=3$$
This point, $$(2,3)$$, satisfies both curves.
At $$(2,3)$$
$$y'_1=4$$
Equation of tangent at $$(2,3)$$
$$(y-3)=4(x-2)$$
$$4x-y-5=0 $$

For normal, slope will be $$-\dfrac{1}{4}$$
$$\therefore$$ Equation of normal at $$(2,3)$$ is
$$(y-3)=\dfrac{-1}{4}(x-2)$$
$$4y-12+x-2=0$$
$$4y+x-14=0$$

Hence, option A.

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Tangents and its Equations Test 8

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Question 1

$$x+y+1=0$$

$$x+y=0$$

$$x-y+1=0$$

$$x-y=0$$

Solution

Question 2

$$y=0 $$

$$x=0$$

$$x+y=0$$

$$x-y=0$$

Solution

Question 3

$$\pm 1$$

$$\pm 2$$

$$\pm 3$$

$$\pm 4$$

Solution

Question 4

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

$$\dfrac{2}{\sqrt{3}}$$

$$\dfrac{8}{\sqrt{3}}$$

$$\dfrac{5}{\sqrt{3}}$$

Solution

Question 5

$$y=mx+m$$

$$y=-mx+2m$$

$$y=m^2x+2m$$

$$y=m^2x+m$$

Solution

Question 6

[-1/2,0]

[-1,-1/2]

[0,1]

[1/2, 1]

Solution

Question 7

$$1, 2$$

$$1, -2$$

$$-1, 2$$

$$-1, -2$$

Solution

Question 8

$$(27, 3)$$

$$(3, 27)$$

$$(3, 3)$$

$$(1, 1)$$

Solution

Question 9

$$2$$ sq.units

$$4$$ sq.units

$$8$$ sq.units

$$16$$ sq.units

Solution

Question 10

tangent as $$4x-y-5=0$$

tangent as $$x+4y-14=0$$

normal as $$4x+y=11$$

normal as $$x-4y=10$$

Solution

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