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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,
 $h = \dfrac{{\sqrt 2 \sec \theta }}{2} \Rightarrow \cos \theta = \dfrac{1}{{\sqrt 2 h}}$ $\ldots(i)$ $k = \dfrac{{\cos {\rm{ec}}\theta }}{2} \Rightarrow \sin \theta = \dfrac{1}{{2k}}$ $\ldots(ii)$
Squaring and adding ($i$) and ($ii$), we get:
 $\dfrac{1}{{2{h^2}}} + \dfrac{1}{{4{k^2}}} = 1$
Therefore, the locus of $R$ is:
 $\dfrac{1}{{2{x^2}}} + \dfrac{1}{{4{y^2}}} = 1$