Tangents and its Equations Test 34

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Tangents and its Equations Test 34

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Weekly Quiz Competition

Question 1
1 / -0

If the curves $$y^2 = 4ax$$ and $$xy = c^2$$ cut orthogonally then $$\dfrac{c^4}{a^4} =$$

A
$$4$$

B
$$8$$

C
$$16$$

D
$$32$$

Solution

$$y^2=4ax, xy=c^2$$ cut orthogonally
Let they intersect at $$(x_1, y_1)$$
$$y^2=4ax$$
$$2y\dfrac{dy}{dx}=4a$$
$$\dfrac{dy}{dx}=\dfrac{2a}{y}$$
$$\dfrac{dy}{dx}|_{(x_1, y_1)}=\dfrac{2a}{y_1}$$ …..$$(1)$$
$$xy=c^2$$
$$y+x\dfrac{dy}{dx}=0$$
$$\dfrac{dy}{dx}|_{(x_1, y_1)}=\dfrac{-y_1}{x_1}$$ ……$$(2)$$
From $$(1), (2)$$ $$(\because m_1m_2=-1)$$
$$\dfrac{2a}{\not{y_1}}\times \dfrac{\not{-}\not{y_1}}{x_1}=\not{-}1$$
$$x_1=2a$$
From $$y^2=4ax$$
$$y_1=\sqrt{8a^2}=2\sqrt{2}a$$
$$xy=c^2$$
$$x_1y_1=c^2$$
$$2a(2\sqrt{2}a)=c^2$$
$$\dfrac{c^2}{a^2}=4\sqrt{2}$$
$$\dfrac{c^4}{a^4}=32$$.
Question 2
1 / -0

The equation of the tangent to the curve $$y = b{e^{ -\dfrac{x}{a}}}$$ at a point , where $$x=0$$ is

A
$$\dfrac{x}{a} - \dfrac{y}{b} = 1$$

B
$$\dfrac{y}{b} - \dfrac{x}{a} = 1$$

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

D
$$\dfrac{x}{b} + \dfrac{y}{a} = 1$$

Solution

$$y=b{ e }^{ \cfrac { -x }{ a } }$$ when $$x=0\, y=b$$
$$\dfrac { dy }{ dx } =\cfrac { -b }{ a } { e }^{ \cfrac { -x }{ a } }$$ at $$x=0$$,
$$\dfrac { dy }{ dx } =\cfrac { -b }{ a } $$
$$y=\cfrac { -bx }{ a } +c$$
$$y=\cfrac { -bx }{ a } +b\\ \cfrac { y }{ b } =\cfrac { -x }{ a } +1\\ \cfrac { x }{ a } +\cfrac { y }{ b } =1$$
Question 3
1 / -0

The equation of the normal to the curve $$y=x+\sin x \cos x$$ at $$x=\pi/2$$ is

A
$$x=2$$

B
$$x=\pi$$

C
$$x+\pi =0$$

D
$$2x=\pi $$

Solution

$$y=x+\sin x\cos x = x+\dfrac{\sin(2x)}{2}$$
$$y'=1+\cos(2x)$$
Substituting $$x=\pi/2$$
$$y'(\pi/2) = 1+\cos(\pi)=0$$

Therefore, slope of tangent is 0. This means that the normal is a vertical line of the $$x=c$$. Since the point has x-coordinate $$\pi/2$$, the equation of the line is:
$$x=\pi/2\implies\boxed{2x=\pi}$$
Question 4
1 / -0

The equation of the normal to the curve $$y=\sin x$$ at $$(0,0)$$ is

A
$$x=0$$

B
$$y=0$$

C
$$x+y=0$$

D
$$x-y=0$$

Solution

$$\textbf{Step-1: Finding the slope of normal at the given point}$$
$$\text{We know that the slope of Tangent means the derivative of that equation.}$$
$$\text{Hence, the slope of the tangent}$$ $$y=sinx$$ $$\text{will be}$$
$$\dfrac{dy}{dx}=cosx$$
$$\text{Also, we know that product of slopes of tangent and normal is}$$ $$-1$$
$$\text{Hence, slope of the normal is}$$ $$\dfrac{-1}{cosx}$$
$$\text{At}$$ $$x=0,$$ $$\text{The slope is}$$ $$-1$$

$$\textbf{Step-2: Finding the equation of the normal}$$
$$\text{Using the general form}$$ $$y-y_1=m(x-x_1)$$ $$\text{we get}$$
$$y-0=-1(x-0)$$
$$x+y=0$$

$$\textbf{Hence, equation of the normal is}$$ $$\mathbf{x+y=0}$$
Question 5
1 / -0

The equation of the curve passing through $$(1,3)$$ whose slope at any point $$(x,y)$$ on it is $$\dfrac { y }{ { x }^{ 2 } }$$ is given by

A
$$y={ 3e }^{ -1/x }$$

B
$$y={ 3e }^{ 1-1/x }$$

C
$$y={ ce }^{ 1/x }$$

D
$$y={ 3e }^{ 1/x }$$

Solution

$$\dfrac{{dy}}{{dx}} = \dfrac{y}{{{x^2}}}\,\,\,\,\,\,\,\left( {given} \right)$$

$$\int {\dfrac{{dy}}{{dx}} = \int {\dfrac{{dx}}{{{x^2}}}} } $$

$$ \Rightarrow \log y = - \dfrac{1}{x} + c$$

$$ \Rightarrow y = {e^{ - \frac{1}{x} + c}}$$

$$ \Rightarrow y = {e^{ - \frac{1}{x} + c}}$$

$$ \Rightarrow y = {e^{ - \frac{1}{x}}},c$$

since this curve is passing through point $$\left( {1,3} \right)$$

so, $$3 = {e^{ - \dfrac{1}{t}}}.c$$

$$ \Rightarrow c = 3e$$

therefore , eqn of the curve is

$$y = {e^{ - \dfrac{1}{x}}}.3e$$

$$ \Rightarrow y = 3{e^{ - \dfrac{1}{x} + 1}}$$

$$ \Rightarrow y = 3{e^{1 - \dfrac{1}{x}}}$$
Question 6
1 / -0

An equation of the tangent to the curve $$y=x^{4}$$ from the point $$(2,0)$$ not on the curve is:

A
$$y=0$$

B
$$x=0$$

C
$$x+y=0$$

D
$$none\ of\ these$$

Solution

let a,b be point on curve (x, y)

$$ \displaystyle b = a^4 $$

$$ \displaystyle \frac{dy}{dx} (x^4) = 4x^3 = 4a^3 $$

$$ \displaystyle (y- y_1) = m (x - x_1) \rightarrow$$ be tangent equation passes through (2, 0)

$$ \displaystyle (b-0) = 4a^3 (a-2) $$

$$ \displaystyle a^4 = 4a^3 (a-2) $$

$$ \displaystyle 3a^4 - 8a^3 = 0 $$

$$ \displaystyle a^3 (3a - 8) = 0 $$

$$ \displaystyle a= 0 \quad or \quad a= \frac{8}{3} \quad (0,0)$$ & $$\left[\dfrac{8}{3},\left(\dfrac{8}{3} \right)^4\right] $$ points on curve

$$ \displaystyle b=0 \quad or \quad b= \left(\frac{8}{3}\right)^4 $$

$$ \displaystyle (y - 0) = \frac{0-0}{2-0} (x-0)\Rightarrow y = 0 \rightarrow $$tangent (1)

$$ \displaystyle \left[y-\left(\frac{8}{3}\right)^4\right] = \frac{(\frac{8}{3})^4_{-0}}{(\frac{8}{3})_{-2}} \quad (x - \frac{8}{3}) \rightarrow $$tangent (2)
Question 7
1 / -0

Find the points on the ellipse $$\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{9}=1$$ , on which the normals are parallel to the line $$3x-y=1$$.

A
$$(\pm\dfrac{6}{\sqrt {10}},\pm\dfrac{3}{\sqrt {10}})$$

B
$$(\pm\dfrac{3}{\sqrt {10}},\pm\dfrac{1}{\sqrt {10}})$$

C
$$(\pm\dfrac{1}{\sqrt {10}},\pm\dfrac{2}{\sqrt {10}})$$

D
None of these

Solution
Question 8
1 / -0

Number of different points on the curve $$y^2=x(x+1)^2$$ where the tangent to the curve drawn at $$(1, 2)$$ meets the curve, is?

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution
Question 9
1 / -0

Find the slope of the normal to the curve $$2x^{2} - xy + 3y^{2} = 18$$ at $$(3,1)$$.

A
$$\dfrac {-11}{3}$$

B
$$\dfrac {3}{11}$$

C
$$\dfrac {11}{3}$$

D
$$\dfrac {-3}{11}$$

Solution

Equation of curve $$2{ x }^{ 2 }-xy+3{ y }^{ 2 }=18$$

Differentiating above equation w.r.t $$x,$$

$$4x-y-x\dfrac { dy }{ dx }+6y\dfrac { dy }{ dx } =0$$

$$4x-y+\dfrac { dy }{ dx } (6y-x)=0$$

$$\dfrac { dy }{ dx } (6y-x)=y-4x$$

$$\dfrac { dy }{ dx } =\dfrac { y-4x }{ 6y-x } $$

$$\dfrac { dy }{ dx| } _{ (3,1) }=\dfrac { 1-4\left( 3 \right) }{ 6-3 } $$

$$\dfrac { dy }{ dx| } _{ (3,1) }=\dfrac { -11 }{ 3 } $$

Slope of normal to the curve
$$=\dfrac { -1 }{ \dfrac { dy }{ dx| } _{ (3,1) } } $$

$$=\dfrac { -1 }{ \dfrac { -11 }{ 3 } } $$

$$=\dfrac { 3 }{ 11 } $$

$$\boxed { \therefore Ans =\dfrac { 3 }{ 11 } } $$
Question 10
1 / -0

Area of the triangle formed by the tangent at $$x=2$$ on the curve $$y= \dfrac{8}{4+x^2}$$ with the coordinate axes is (in sq. units)

A
$$4$$

B
$$3$$

C
$$2$$

D
$$1$$

Solution

$$y=\dfrac {8}{4+x^2}$$

$$At \,\,x=2\ y=1$$

$$y' =\dfrac {-8(2x)}{(4+x^2)^2}=\dfrac {-16\times 2}{8^2}=\dfrac {-1}{2}$$

Tangent is $$y=mx+c$$

$$y=\dfrac {-1}{2}x+c$$

passes through $$(2, 1)\ C=2$$

$$2y+x=4$$
Base of triangle $$=4$$

Height of triangle $$=\dfrac {4}{2}=2$$

Area of $$\Delta =\dfrac {1}{2}\times 4\times 2=4\ sq.unit$$

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Tangents and its Equations Test 34

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Tangents and its Equations Test 34

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SHARING IS CARING

Question 1

$$4$$

$$8$$

$$16$$

$$32$$

Solution

Question 2

$$\dfrac{x}{a} - \dfrac{y}{b} = 1$$

$$\dfrac{y}{b} - \dfrac{x}{a} = 1$$

$$\dfrac{x}{a} + \dfrac{y}{b} = 1$$

$$\dfrac{x}{b} + \dfrac{y}{a} = 1$$

Solution

Question 3

$$x=2$$

$$x=\pi$$

$$x+\pi =0$$

$$2x=\pi $$

Solution

Question 4

$$x=0$$

$$y=0$$

$$x+y=0$$

$$x-y=0$$

Solution

Question 5

$$y={ 3e }^{ -1/x }$$

$$y={ 3e }^{ 1-1/x }$$

$$y={ ce }^{ 1/x }$$

$$y={ 3e }^{ 1/x }$$

Solution

Question 6

$$y=0$$

$$x=0$$

$$x+y=0$$

$$none\ of\ these$$

Solution

Question 7

$$(\pm\dfrac{6}{\sqrt {10}},\pm\dfrac{3}{\sqrt {10}})$$

$$(\pm\dfrac{3}{\sqrt {10}},\pm\dfrac{1}{\sqrt {10}})$$

$$(\pm\dfrac{1}{\sqrt {10}},\pm\dfrac{2}{\sqrt {10}})$$

None of these

Solution

Question 8

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 9

$$\dfrac {-11}{3}$$

$$\dfrac {3}{11}$$

$$\dfrac {11}{3}$$

$$\dfrac {-3}{11}$$

Solution

Question 10

$$4$$

$$3$$

$$2$$

$$1$$

Solution

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