Answer to Question #269032 in Analytic Geometry for Faru

Question

If l, m be the portions of the axes of x and y intercepted by the tangent at any point ( x,y) on the curve ( x/a )^2/3+  (  y/b)^2/3  , show that ( l^2/a^2) +(m^2/b^2)=1

Answer 1

l, m be the portions of the axes of x and y intercepted by the tangent at any point ( x,y) on the curve

$( \frac{x}{a} )^{\frac{2}{3}}+ (\frac{y}{b})^{\frac{2}{3}}=1$

$b^{\frac{2}{3}}x^{\frac{2}{3}}+a^{\frac{2}{3}}y^{\frac{2}{3}}=a^{\frac{2}{3}}b^{\frac{2}{3}}$ ••••equation(1)

The parametric coordinate of the point (x,y) is ($acos³\theta , bsin³\theta)$ , $\theta$ is a parameter.

By the problem, equation of tangent is

$\frac{x}{l}+\frac{y}{m}=1$ ••••equation(2) to the curve at ( acos³$\theta$ , bsin³$\theta)$

Now we find equation of tangent at (acos³$\theta$ , bsin³$\theta)$

Differentiating equation (1) with respect to x we get

$\frac{2}{3}b^{\frac{2}{3}}x^{-\frac{1}{3}} + \frac{2}{3}a^{\frac{2}{3}}y^{-\frac{1}{3}} \frac{dy}{dx}=0$

=> $\frac{dy}{dx} = -\frac{ {b^{\frac{2}{3}}y^{\frac{1}{3}}}}{ {a^{\frac{2}{3}}x^{\frac{1}{3}}}}$

Now $[\frac{dy}{dx}]_{(acos³\theta , bsin³\theta)}= -\frac{ {b^{\frac{2}{3}}(bsin³\theta)^{\frac{1}{3}}}}{ {a^{\frac{2}{3}}(acos³\theta)^ {\frac{1}{3}}}} =-\frac{b}{a}tan\theta$

So equation of tangent at $(acos³\theta , bsin³\theta)$ will be

$y-bsin³\theta$ =$-\frac{b}{a}tan\theta(x-acos³\theta)$

=> $(acos\theta)y + (bsin\theta)x = absin\theta cos³\theta + abcos\theta sin³\theta$

=> $(bsin\theta)x +(acos\theta)y = absin\theta cos\theta (cos²\theta + sin²\theta)$

=> $(bsin\theta)x +(acos\theta)y = absin\theta cos\theta$

=> $\frac{(bsin\theta)x +(acos\theta)y}{absin\theta cos\theta }= 1$

=> $\frac{x}{acos\theta}+\frac{y}{bsin\theta}=1$ ••••equation(3)

Obviously equation (2) and equation (3) are identical.

So equating coefficients we get

$\frac{l}{acos\theta}=\frac{m}{bsin\theta}=\frac{1}{1}$

=> l = acos$\theta$ and m = bsin$\theta$

=> $\frac{l}{a} = cos\theta$ and $\frac{m}{b} = sin\theta$

So ${(\frac{l}{a})}^{2} +{(\frac{m}{b})}^{2}= cos²\theta + sin²\theta$

=> ${(\frac{l}{a})}^{2} +{(\frac{m}{b})}^{2}= 1$

=> ${(\frac{l²}{a²})}+{(\frac{m²}{b²})}=1$

Hence proved.