Tangents and its Equations Test 55

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Tangents and its Equations Test 55

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Question 1
1 / -0

Paraboals $$(y-\alpha )^{2}=4a(x-\beta )and (y-\alpha )^{2}=4a'(x-\beta ')$$ will have a common normal (other than the normal passing through vertex ofparabola)if:

A
$$\frac{2(a-a')}{\beta '-\beta }< 1$$

B
$$\frac{2(a-a')}{\beta -\beta' }< 1$$

C
$$\frac{2(a'-a)}{\beta -\beta' }< 1$$

D
$$\frac{2(a'a)}{\beta -\beta' }> 1$$

Solution
Question 2
1 / -0

The normal to the curve $$x=a\left( \cos { \theta } +\theta \sin { \theta } \right) $$, $$y=a\left( \sin { \theta } -\theta \cos { \theta } \right) $$ at any point $$\theta$$ is such that

A
it passes through origin

B
it passes through the point $$(1,1)$$

C
it passes through $$\left( \frac { a\pi }{ 2 } ,-a \right) $$

D
it is at a constant distance from the origin

Solution

$$x=a\cos \theta +a\theta \sin \theta $$
$$\dfrac {dx}{d\theta }=-a\sin \theta +a\left\{\theta (-\cos \theta )+\sin \theta \right\}$$
$$=-a\theta \cos \theta $$
$$y=a(\sin \theta -\theta \cos \theta )$$
$$\dfrac {dy}{dx}=a\cos \theta -a \left\{\theta \sin \theta +\cos \theta \right\}$$
$$\therefore \ \dfrac {dy}{dx}=\dfrac {-a\theta \sin \theta }{-a\theta \cos \theta }=\tan \theta $$
The equation of normal at $$\theta $$ is:
$$y-y_1=-\dfrac {1}{\dfrac {dy}{dx}} (x-x_1)$$
or, $$y-a (\sin \theta -\theta \cos \theta )=-\dfrac {\cos \theta }{\sin \theta }(x-a\cos \theta -a\theta \sin \theta )$$
or, $$y \sin \theta -\sin^2\theta +a\theta \sin \theta \cos \theta $$
$$=-x\cos \theta +a\cos^2\theta +a\theta \sin \theta \cos \theta $$
or, $$y\sin \theta +x\cos \theta =a$$
which wakes it clear that normal passes through origin $$(0,0)$$
Question 3
1 / -0

The slope of the curve $$y=\sin { x } +\cos ^{ 2 }{ x }$$ is zero at a point , whose x-coordinate can be ?

A
$$\dfrac { \pi }{ 4 } $$

B
$$\dfrac { \pi }{ 2 } $$

C
$${ \pi }$$

D
$$\dfrac { \pi }{ 3 } $$

Solution
Question 4
1 / -0

The tangent to the circle $$x^2+y^2=5$$ at the point $$(1,-2)$$ also touches the circle $$x^2+y^2-8x+6y+20=0$$ at

A
$$(-2,1)$$

B
$$(-3,0)$$

C
$$(-1,-1)$$

D
$$(3,-1)$$

Solution
Question 5
1 / -0

If for a curve represented parametrically by $$x={ sec }^{ 2 }t,\quad y=cot\quad t\quad $$ , the tangent at a point $$P(t=\frac { \pi }{ 4 } )$$ meets the curve again at the point Q, then $$\begin{vmatrix} PQ \end{vmatrix}$$is equal to

A
$$\frac { 2\sqrt { 5 } }{ 3 } $$

B
$$\frac { 3\sqrt { 5 } }{ 2 } $$

C
$$\frac { 5\sqrt { 3 } }{ 3 } $$

D
$$\frac { 5\sqrt { 5 } }{ 4 } $$

Solution
Question 6
1 / -0

If the subnormal to the curve $${ x }^{ 2 }.{ y }^{ n }={ a }^{ 2 }$$ is a constant then n=

A
$$-4$$

B
$$-3$$

C
$$-2$$

D
$$-1$$

Solution

Given,

$$x^2y^n=a^2$$

$$\Rightarrow x=\sqrt{\dfrac{a^2}{y^n}}$$

$$\Rightarrow 2xy^n+x^2ny^{n-1}\dfrac{dy}{dx}=0$$

$$\Rightarrow \dfrac{dy}{dx}=-\dfrac{2y}{nx}$$

$$\Rightarrow \dfrac{dy}{dx}=-\dfrac{2y}{n\sqrt{\frac{a^2}{y^n}}}$$

$$\therefore \dfrac{dy}{dx}=-\dfrac{2}{na}y^{\frac{n+2}{2}}$$

Subnormal is given by,

$$SN=y \dfrac{dy}{dx}$$

$$=y\left ( -\dfrac{2}{na}y^{\frac{n+2}{2}} \right )$$

$$\therefore SN=-\dfrac{2}{na}y^{\frac{n+4}{2}}$$

$$\therefore n=-4$$
Question 7
1 / -0

If $$\theta$$ is angle of intersection between $$y=10-x^{2}$$ and $$y=4+x^{2}$$ then $$|\tan \theta|$$ is-

A
$$\dfrac {5\sqrt {3}}{11}$$

B
$$\dfrac {7\sqrt {3}}{15}$$

C
$$\dfrac {4\sqrt {3}}{11}$$

D
$$none$$

Solution

Given:$$y=10-{x}^{2}$$ ......$$(1)$$
$$y=4+{x}^{2}$$ ......$$(2)$$
On solving,
$$10-{x}^{2}=4+{x}^{2}$$
$$\Rightarrow 2{x}^{2}=6$$
$$\Rightarrow {x}^{2}=\dfrac{6}{2}=3$$
$$\therefore x=\pm \sqrt{3}$$ from $$(1)$$
$$\Rightarrow y=10-{\left(\pm\sqrt{3}\right)}^{2}=10-3=7$$
Consider $$y=10-{x}^{2}$$
$$\dfrac{dy}{dx}=-2x$$
$${m}_{1}=\left[\dfrac{dy}{dx}\right]_{\left(\sqrt{3},7\right)}=-2\times\sqrt{3}=-2\sqrt{3}$$
Consider $$y=4+{x}^{2}$$
$$\dfrac{dy}{dx}=2x$$
$${m}_{2}=\left[\dfrac{dy}{dx}\right]_{\left(\sqrt{3},7\right)}=2\times\sqrt{3}=2\sqrt{3}$$
$$=\left|\dfrac{{m}_{1}-{m}_{2}}{1+{m}_{1}{m}_{2}}\right|$$
$$\tan{\theta}=\left|\dfrac{-2\sqrt{3}-2\sqrt{3}}{1+\left(-2\sqrt{3}\right)\left(2\sqrt{3}\right)}\right|$$
$$=\left|\dfrac{-4\sqrt{3}}{1-12}\right|$$
$$=\dfrac{4\sqrt{3}}{11}$$
$$\therefore \tan{\theta}=\dfrac{4\sqrt{3}}{11}$$
Question 8
1 / -0

$$y = 6\tan \,x\left( {{e^x} - x - 1} \right) - 3{x^3} - {x^4} - \frac{5}{4}{x^5},\,$$ if $${n^{th}}$$ derivative at x=0 is non zero then least value of n is

A
3

B
4

C
5

D
6

Solution
Question 9
1 / -0

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

A
$$(1,1)$$

B
At no point

C
$$(0,1)$$

D
$$(1,0)$$

Solution
Question 10
1 / -0

The slope of the tangent to the curve at a point (x, y) on it is proportional to (x-2). If the slope of the tangent to the curve at (10,-9) on it -3. The equation of the curve is

A
$$y=k{ \left( x-2 \right) }^{ 2 }$$

B
$$y=\frac { -3 }{ 16 } { \left( x-2 \right) }^{ 2 }\ +1$$

C
$$y=\frac { -3 }{ 16 } { \left( x-2 \right) }^{ 2 }\ +3$$

D
$$y=k{ \left( x+2 \right) }^{ 2 }$$

Solution

$${\textbf{Step -1: Find the value of slope}}{\textbf{.}}$$
$${\text{The slope of the tangent to the curve at a point }}\left( {x,y} \right){\text{ on it is proportional to }}\left( {x - 2} \right).$$
$$ = \dfrac{{dy}}{{dx}} = m\left( {x - 2} \right)$$
$$ = \dfrac{{dy}}{{dx}} = k\left( {x - 2} \right)\left[ {{\text{where }}k{\text{ is a proportionality constant}}} \right]$$
$${\text{Slope of the tangent to the curve at }}\left( {10, - 9} \right){\text{ is }} - 3.$$
$$ \Rightarrow - 3 = k\left( {10 - 2} \right)$$
$$ \Rightarrow - 3 = 8k$$
$$ \Rightarrow k = - \dfrac{3}{8}$$
$$ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{3}{8}\left( {x - 2} \right)$$
$${\textbf{Step -2: Integrate the formed equation and solve it further}}{\textbf{.}}$$
$${\text{Integrating both sides,}}$$
$$ \Rightarrow y = - \dfrac{3}{8}\dfrac{{{{\left( {x - 2} \right)}^2}}}{2} + c{\text{ }}.......\left( i \right){\text{ }}\left[ {{\text{Eqaution of curve}}} \right]$$
$${\text{Also, }}\left( {10, - 9} \right){\text{ satisfy the equation of curve as it lies on the curve}}{\text{.}}$$
$$\therefore - 9 = - \dfrac{3}{8}\dfrac{{{{\left( {10 - 2} \right)}^2}}}{2} + c$$
$$ \Rightarrow - 9 = - \dfrac{3}{{16}} \times {8^2} + c$$
$$ \Rightarrow c = 3$$
$${\textbf{Hence, the equation of the curve is }}\mathbf{y= - \dfrac{3}{{16}}{\left( {x - 2} \right)^2} + 3.}$$