ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 9- Worked Out Examples

Saturday 2 August 2014

CHAPTER 9- Worked Out Examples

Example: 1

Prove that the segment of the tangent to $xy = {c^2}$ intercepted between the axes is bisected at the point of contact.

Solution: 1

Let us take an arbitrary point on this curve, $\left( {t,\dfrac{{{c^2}}}{t}} \right)$ . An approximate figure showing the tangent at this point is sketched below:

The procedure that we need to follow is first determine the equation of the tangent at the point $P$ , find the intercepts this tangent makes with the axes (we will then get the co-ordinates of the points $A$ and $B$ ), and show that $P$ is the mid-point of $AB$ .

${m_T}\left( {{\rm{at}}\,P} \right) = {\left. {\dfrac{{dy}}{{dx}}} \right|_{x = t}} = {\left. {\dfrac{{ - {c^2}}}{{{x^2}}}} \right|_{x = t}} = \dfrac{{ - {c^2}}}{{{t^2}}}$

Equation of tangent: $\,\,\,$ $y - \dfrac{{{c^2}}}{t} = \dfrac{{ - {c^2}}}{{{t^2}}}\left( {x - t} \right)$

Point A: $\,\,\,$ Put $\,\,$ $x=0$

	$\Rightarrow \,\,\, y = \dfrac{{{c^2}}}{t} + \dfrac{{{c^2}}}{t} = \dfrac{{2{c^2}}}{t}$
	$\Rightarrow$ The point $A$ is $\left( {0,\dfrac{{2{c^2}}}{t}} \right)$

Point B: $\,\,\,$ Put $\,\,$ $y=0$

	$\Rightarrow \,\,\,x = t + t = 2t$
	$\Rightarrow$ The point $B$ is $(2t,0)$

Mid-point of AB: $\,\,\,$ The mid-point of AB is

	$\left( {\dfrac{{0 + 2t}}{2},\dfrac{{\dfrac{{2{c^2}}}{t} + 0}}{2}} \right)$
	$= \left( {t,\dfrac{{{c^2}}}{t}} \right)$
	which is the same as the point $P$ .
	$\Rightarrow \,\, P$ is the mid-point of $AB$

Example: 2

Find the equations of tangents to the curve $y = {x^4}$ which are drawn from the point $(2,0)$ .

Solution: 2

We write the equation of the tangent to $y = {x^4}$ at a general point $(t,{t^4})$ and then make $(2,0)$ satisfy that equation.

${m_T} = {\left. {\dfrac{{dy}}{{dx}}} \right|_{x = t}} = {\left. {4{x^3}} \right|_{x = t}} = 4{t^3}$

Equation of tangent: $\,\,\,$ $y - {t^4} = 4{t^3}\left( {x - t} \right)$

$\Rightarrow \,\,\, 4{t^3}x - y - 3{t^4} = 0$

$\ldots(i)$

Since the tangent we require passes from $(2,0)$ , the co-ordinates of $(2,0)$ must satisfy ( $i$ )

	$\Rightarrow \,\,\, 4{t^3}\left( 2 \right) - \left( 0 \right) - 3{t^4} = 0$
	$\Rightarrow \,\,\, 8{t^3} - 3{t^4} = 0$
	$\Rightarrow \,\,\, {t^3}\left( {8 - 3t} \right) = 0$
	$\Rightarrow \,\,\, t = 0,\dfrac{8}{3}$

From( $i$ ), the two possible tangents are (corresponding to the two values of $t$ ):

$y = 0;\,\,\,\,\,\,y - {\left( {\dfrac{8}{3}} \right)^4} = 4{\left( {\dfrac{8}{3}} \right)^3}\left( {x - \dfrac{8}{3}} \right)$

Example: 3

Find the point(s) on the curve ${y^3} + 3{x^2} = 12y$ where the tangent is vertical.

Solution: 3

A vertical tangent means that the slope of the tangent is $\infty$ .

Differentiating the equation of the given curve w.r.t $x$ , we get:

	$3{y^2}\dfrac{{dy}}{{dx}} + 6x = 12\dfrac{{dy}}{{dx}}$
	$\Rightarrow \,\,\, \dfrac{{dy}}{{dx}} = \dfrac{{6x}}{{12 - 3{y^2}}}$

Hence, for vertical tangent:

	$3{y^2} = 12$
	$\Rightarrow \,\,\, y = \pm \,2$
	$\Rightarrow \,\,\, 3{x^2} = y\left( {12 - {y^2}} \right)$
	$=\pm \,\,16$
	$\Rightarrow \,\,\, x = \pm \dfrac{4}{{\sqrt 3 }}\left( {{\rm{for}}\,\,y = 2} \right)\,\,\,\,{{\rm{for\,y = - 2\,x \,has\, imaginary\, values}}}$

Therefore, the required points are $\left( { \pm \dfrac{4}{{\sqrt 3 }},2} \right)$

Example: 4

Tangents are drawn to the ellipse ${x^2} + 2{y^2} = 2$ . Find the locus of the mid-point of the intercept made by the tangent between the co-ordinate axes.

Solution: 4

To determine the required locus, we first write down the equation of an arbitrary tangent to the given ellipse:

$\dfrac{{{x^2}}}{2} + {y^2} = 1$

A general point on this ellipse can be taken as $\left( {\sqrt 2 \cos \theta ,\sin \theta } \right)$ . Now we write the equation of the tangent at this point by first differentiating the equation of the ellipse:

	$x + 2y\dfrac{{dy}}{{dx}} = 0$
	$\Rightarrow \,\,\, \dfrac{{dy}}{{dx}} = \dfrac{{ - x}}{{2y}}$
	$\Rightarrow \,\,\, {m_T}\left( {{\rm{at}}\left( {\sqrt 2 \cos \theta ,\sin \theta } \right)} \right) = \dfrac{{ - \sqrt 2 \cos \theta }}{{2\sin \theta }} = \dfrac{{ - \cot \theta }}{{\sqrt 2 }}$

Equation of tangent: $\,\,\,$ $y - \sin \theta = \dfrac{{ - \cot \theta }}{{\sqrt 2 }}\left( {x - \sqrt 2 \cos \theta } \right)$

$\Rightarrow \,\,\, x\cos \theta + \sqrt 2 y\sin \theta = \sqrt 2$

$x$ -intercept $\,\,\,$ Put $\,\,$ $y = 0 \,\,\, \Rightarrow \;\;\;\;\;x = \sqrt 2 \sec \theta$

$\Rightarrow\,\,\,$

The tangent intersects the $x$ -axis at $P\left( {\sqrt 2 \sec \theta ,0} \right)$

$y$ -intercept: $\,\,\,$ Put $\,\,x = 0 \,\,\, \Rightarrow \;\;\;\;\;y = \cos ec\theta$

$\Rightarrow\,\,\,$

The tangent intersects the $y$ -axis at $Q(0,\cos ec\theta )$

We require the locus of $R$ , the mid-point of $PQ$ . Let its co–ordinates be $(h,k)$ . Therefore,

	$h = \dfrac{{\sqrt 2 \sec \theta }}{2} \Rightarrow \cos \theta = \dfrac{1}{{\sqrt 2 h}}$	$\ldots(i)$
	$k = \dfrac{{\cos {\rm{ec}}\theta }}{2} \Rightarrow \sin \theta = \dfrac{1}{{2k}}$	$\ldots(ii)$