tg

tg
tgt

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
Post a Comment