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=0x+y+1=0x+y+1=0

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

x−y+1=0x-y+1=0x−y+1=0

x−y=0x-y=0x−y=0

Solution

Question 2

y=0y=0 y=0

x=0x=0x=0

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

x−y=0x-y=0x−y=0

Solution

Question 3

±1\pm 1±1

±2\pm 2±2

±3\pm 3±3

±4\pm 4±4

Solution

Question 4

43\dfrac{4}{\sqrt{3}}3​4​

23\dfrac{2}{\sqrt{3}}3​2​

83\dfrac{8}{\sqrt{3}}3​8​

53\dfrac{5}{\sqrt{3}}3​5​

Solution

Question 5

y=mx+my=mx+my=mx+m

y=−mx+2my=-mx+2my=−mx+2m

y=m2x+2my=m^2x+2my=m2x+2m

y=m2x+my=m^2x+my=m2x+m

Solution

Question 6

[-1/2,0]

[-1,-1/2]

[0,1]

[1/2, 1]

Solution

Question 7

1,21, 21,2

1,−21, -21,−2

−1,2-1, 2−1,2

−1,−2-1, -2−1,−2

Solution

Question 8

(27,3)(27, 3)(27,3)

(3,27)(3, 27)(3,27)

(3,3)(3, 3)(3,3)

(1,1)(1, 1)(1,1)

Solution

Question 9

222 sq.units

444 sq.units

888 sq.units

161616 sq.units

Solution

Question 10

tangent as 4x−y−5=04x-y-5=04x−y−5=0

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

normal as 4x+y=114x+y=114x+y=11

normal as x−4y=10x-4y=10x−4y=10

Solution

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$x+y+1=0$

$x+y=0$

$x-y+1=0$

$x-y=0$

$y=0$

$x=0$

$x+y=0$

$x-y=0$

$\pm 1$

$\pm 2$

$\pm 3$

$\pm 4$

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

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

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

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

$y=mx+m$

$y=-mx+2m$

$y=m^2x+2m$

$y=m^2x+m$

$1, 2$

$1, -2$

$-1, 2$

$-1, -2$

$(27, 3)$

$(3, 27)$

$(3, 3)$

$(1, 1)$

$2$ sq.units

$4$ sq.units

$8$ sq.units

$16$ sq.units

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

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

normal as $4x+y=11$

normal as $x-4y=10$