Tangents and its Equations Test 2

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Tangents and its Equations Test 2

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Weekly Quiz Competition

Question 1
1 / -0

Let $$C$$ be a curve given by $$y(x) = 1 + \sqrt {4x - 3}, x > \dfrac {3}{4}$$. If $$P$$ is a point on $$C$$, such that the tangent at $$P$$ has slope $$\dfrac {2}{3}$$, then a point through which the normal at $$P$$ passes, is:

A
$$(1, 7)$$

B
$$(3, -4)$$

C
$$(4, -3)$$

D
$$(2, 3)$$

Solution

$$\dfrac {dy}{dx} = \dfrac {1}{2\sqrt {4x - 3}} \times 4 = \dfrac {2}{3}$$
$$\Rightarrow 4x - 3 = 9$$
$$\Rightarrow x = 3$$
So, $$y = 4$$
Equation of normal at $$P(3, 4)$$ is
$$y - 4 = -\dfrac {3}{2} (x - 3)$$
i.e. $$2y - 8 = -3x + 9$$
$$\Rightarrow 3x + 2y - 17 = 0$$
Question 2
1 / -0

What is the $$x$$-coordinate of the point on the curve $$f(x) = \sqrt {x}(7x - 6)$$, where the tangent is parallel to $$x$$-axis?

A
$$-\dfrac {1}{3}$$

B
$$\dfrac {2}{7}$$

C
$$\dfrac {6}{7}$$

D
$$\dfrac {1}{2}$$

Solution

Given, $$f(x) = \sqrt {x}(7x - 6) = 7x^{3/2} - 6x^{1/2}$$
Thus $$f'(x) = 7\times \dfrac {3}{2}x^{1/2} - 6\times \dfrac {1}{2} x^{-1/2}$$
When tangent is parallel to $$x$$ axis $$f'(x) = 0$$.
Therefore, $$\dfrac {21}{2}x^{1/2} - 3x^{-1/2} = 0$$
$$\Rightarrow \dfrac {21}{2}\sqrt {x} = \dfrac {3}{\sqrt {x}}$$
$$\Rightarrow 7x = 2$$
$$\Rightarrow x = \dfrac {2}{7}$$
Question 3
1 / -0

Tangents are drawn from the origin to the curve $$y=\cos { x } $$. Their points of contact lie on

A
$${ x }^{ 2 }{ y }^{ 2 }={ y }^{ 2 }-{ x }^{ 2 }$$

B
$${ x }^{ 2 }{ y }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }$$

C
$${ x }^{ 2 }{ y }^{ 2 }={ x }^{ 2 }-{ y }^{ 2 }$$

D
None of these

Solution

Let $$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ be one of the points of contact. Given curve is $$y=\cos { x } $$
$$\Rightarrow \cfrac { dy }{ dx } =-\sin { x } $$
$$\Rightarrow { \left| \cfrac { dy }{ dx } \right| }_{ \left( { x }_{ 1 },{ y }_{ 1 } \right) }^{ }=-\sin { { x }_{ 1 } } $$
Now the equation of the tangent at $$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ is
$$y-{ y }_{ 1 }{ \left( \cfrac { dy }{ dx } \right) }_{ \left( { x }_{ 1 },{ y }_{ 1 } \right) }\left( x-{ x }_{ 1 } \right) $$
$$\Rightarrow y-{ y }_{ 1 }=-\sin { { x }_{ 1 } } \left( 0-{ x }_{ 1 } \right) $$
Since, it is given that equation of tangent passes through origin
$$\Rightarrow 0-{ y }_{ 1 }=-\sin { { x }_{ 1 } } \left( 0-{ x }_{ 1 } \right) $$
$$\therefore { y }_{ 1 }=-{ x }_{ 1 }\sin { { x }_{ 1 } } ...(i)$$
also, point $$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ lies on $$\therefore { y }_{ 1 }=\cos { { x }_{ 1 } } $$
From Eqs (i), (ii) we get
$$\sin ^{ 2 }{ { x }_{ 1 } } +\cos ^{ 2 }{ { x }_{ 1 } } =\cfrac { { y }_{ 1 }^{ 2 } }{ { x }_{ 1 }^{ 2 } } +{ y }_{ 1 }^{ 2 }=1$$
$$ \Rightarrow { x }_{ 1 }^{ 2 }={ y }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }{ x }_{ 1 }^{ 2 }\quad $$
Hence, the locus of $${ x }^{ 2 }={ y }^{ 2 }+{ y }^{ 2 }{ x }^{ 2 }\Rightarrow { x }^{ 2 }{ y }^{ 2 }={ x }^{ 2 }-{ y }^{ 2 }$$
Question 4
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Determine the equation of tangent at vertex of the parabola $$\displaystyle (x+4)^{2}=-4(y-2)$$.

A
$$y=0$$

B
$$y=2$$

C
$$x=0$$

D
$$x+4=0$$

Solution

Given hyperbola is, $$(x+4)^2=-4(y-2)$$

Differentiating w.r.t $$x$$

$$2(x+4)=-4\cfrac{dy}{dx}\Rightarrow \cfrac{dy}{dx}=-\cfrac{x+4}{2}$$

Clearly the vertex of the parabola is, $$(-4,2)$$

Hence slope of the tangent at the vertex is, $$m=\left(\cfrac{dy}{dx}\right)_{(-4,2)}=-\cfrac{-4+4}{2}=0$$

Ergo required tangent is, $$(y-2)=0(x+4)=0$$

$$\Rightarrow y=2$$
Question 5
1 / -0

Normal to the parabola $$\displaystyle y^{2}=4ax$$ where $$m$$ is the slope of the normal is

A
$$\displaystyle y=mx+2am-am^{3}$$

B
$$\displaystyle y=mx-2am-am^{3}$$

C
$$\displaystyle y=mx-2am+am^{3}$$

D
none of these

Solution

Given, $$y^2=4ax$$
$$\Rightarrow 2ydy=4a$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{2a}{y}$$
Thus slope of normal to the given parabola at any point is,
$$\displaystyle =-\left(\frac{1}{\frac{dy}{dx}}\right)=-\frac{y}{2a}=m$$ (given)
$$\Rightarrow y=-2am$$
Thus from $$y^2=4ax,$$ $$x=\dfrac{y^2}{4a}=am^2$$
Therefore the point is $$(am^2,-2am)$$
Hence the required line passing through $$(am^2,-2am)$$ and having slope $$m$$ is,
$$(y+2am)=m(x-am^2)$$
$$\Rightarrow y= mx-2am-am^3$$
Question 6
1 / -0

Tangent to parabola $$\displaystyle y^{2}=4x+5$$ which is parallel to $$y=2x+7$$

A
$$y-2x-3=0$$

B
$$ y=x+3$$

C
$$y-2x+1=0$$

D
$$y=x+1$$

Solution

Given, $$y^2=4x+5$$

$$\therefore \displaystyle 2y\frac{dy}{dx}=4 $$

$$\therefore \dfrac{dy}{dx}=\dfrac{2}{y}=2,$$ (given)

$$\therefore y=1 \Rightarrow x=-1$$

$$\therefore $$ required tangent is given by, $$y-1=2(x+1)$$ or $$y=2x+3$$
Question 7
1 / -0

The slope of tangent to the curve $$y=\int_{0}^{x}\displaystyle \frac{dx}{1+x^{3}}$$ at the point where $$x=1$$ is

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

B
$$1$$

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

D
none of these

Solution

$$y=\int _{ 0 }^{ x }{ \cfrac { dx }{ 1+{ x }^{ 3 } } } \\ \cfrac { dy }{ dx } =\cfrac { 1 }{ 1+{ x }^{ 3 } } =\cfrac { 1 }{ 2 } $$
Question 8
1 / -0

The values of '$$a$$' for which $$y=x^{2}+ ax+25$$ touches $$x$$-axis are

A
$$\pm 10$$

B
$$\pm 2$$

C
$$\pm 1$$

D
0

Solution

$${y}'=2x+a=0$$
For any quadratic equation to touch x-axis it must have equal roots at that point, so the equation must be in form of perfect square.
$$y=(x+5)^{2}$$ at $$a=10$$
Question 9
1 / -0

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

A
$$x-y=0$$

B
$$2x+y=3$$

C
$$x+2y=3$$

D
$$x-y=1$$

Solution

$$\displaystyle 2y=3-x^{2}$$

$$\cfrac{dy}{dx}=-x =m$$

Thus slope of normal at $$(1,1)$$ is $$=-\left(\cfrac{1}{m}\right)_{(1,1)} = 1$$

Hence required normal is, $$(y-1)=1(x-1)\Rightarrow x-y=0$$
Question 10
1 / -0

The equation of the normal to the curve $$\displaystyle x^{3}+y^{3}=6xy$$ at the point $$(3,3).$$

A
$$ x+y-6=0$$

B
$$ -x-y+6=0$$

C
$$ x-y=0$$

D
$$ x+y=0$$

Solution

$$\displaystyle

\dfrac{dy}{dx}=-\dfrac{fx}{fy}=-\dfrac{3x^{2}-6y}{3y^{2}-6x}=-1 $$ at $$(3,3)$$
$$\therefore$$ Required normal is $$y-3=1(x-3)$$ or $$x-y=0$$

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Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 2

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Tangents and its Equations Test 2

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SHARING IS CARING

Question 1

$$(1, 7)$$

$$(3, -4)$$

$$(4, -3)$$

$$(2, 3)$$

Solution

Question 2

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

$$\dfrac {2}{7}$$

$$\dfrac {6}{7}$$

$$\dfrac {1}{2}$$

Solution

Question 3

$${ x }^{ 2 }{ y }^{ 2 }={ y }^{ 2 }-{ x }^{ 2 }$$

$${ x }^{ 2 }{ y }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }$$

$${ x }^{ 2 }{ y }^{ 2 }={ x }^{ 2 }-{ y }^{ 2 }$$

None of these

Solution

Question 4

$$y=0$$

$$y=2$$

$$x=0$$

$$x+4=0$$

Solution

Question 5

$$\displaystyle y=mx+2am-am^{3}$$

$$\displaystyle y=mx-2am-am^{3}$$

$$\displaystyle y=mx-2am+am^{3}$$

none of these

Solution

Question 6

$$y-2x-3=0$$

$$ y=x+3$$

$$y-2x+1=0$$

$$y=x+1$$

Solution

Question 7

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

$$1$$

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

none of these

Solution

Question 8

$$\pm 10$$

$$\pm 2$$

$$\pm 1$$

0

Solution

Question 9

$$x-y=0$$

$$2x+y=3$$

$$x+2y=3$$

$$x-y=1$$

Solution

Question 10

$$ x+y-6=0$$

$$ -x-y+6=0$$

$$ x-y=0$$

$$ x+y=0$$

Solution

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