Answer to Question #268954 in Calculus for Tithi

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=1 , show that (l)^2/(a)^2 + (m)^2/(b)^2 =1

Answer 1

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

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

$(\frac{y}{b}^{\frac{-1}{3}})\times\frac{1}{b})\frac{dy}{dx}=-(\frac{x}{a}^{\frac{-1}{3}})\times\frac{1}{a}$

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

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

$let\ A(x_1,y_1)\implies m=(-\frac{b}{a})^{\frac{2}{3}}(\frac{x}{y})^{\frac{-1}{3}}$

Equation of the tangent is

$y-y_1=m(x-x_1)$

$y-y_1=(-\frac{b}{a})^{\frac{2}{3}}(\frac{x}{y})^{\frac{-1}{3}}(x-x_1)$

open by cross multiplying

$a^{\frac{2}{3}}x_1^{\frac{1}{3}}y-a^{\frac{2}{3}}x^{\frac{1}{3}}y_1=-b^{\frac{2}{3}}y_1^{\frac{1}{3}}x_1+b^{\frac{2}{3}}y_1^{\frac{1}{3}}x_1$

$(b^{\frac{2}{3}}y_1^{\frac{1}{3}})x+(a^{\frac{2}{3}}x^{\frac{1}{3}})y=x_1^{\frac{1}{3}}y_1^{\frac{1}{3}}[a^{\frac{2}{3}}y_1^{\frac{2}{3}}+b^{\frac{2}{3}}x^{\frac{2}{3}}]$

$x_1^{\frac{1}{3}}y_1^{\frac{2}{3}}a^{\frac{2}{3}}b^{\frac{2}{3}}[\frac{x_1}{a}^{\frac{2}{3}}+\frac{y_1}{b}^{\frac{2}{3}}]..................equqation \ 1$

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

equation 1$\implies(b^{\frac{2}{3}}y_1^{\frac{1}{3}})x+(a^{\frac{2}{3}}x^{\frac{1}{3}})y=x_1^{\frac{1}{3}}y_1^{\frac{2}{3}}a^{\frac{2}{3}}b^{\frac{2}{3}}$

$\frac{x}{x_1^{\frac{1}{3}}a^{\frac{2}{3}}}+\frac{y}{y_1^{\frac{1}{3}}b^{\frac{2}{3}}}=1$

as per given a=I b=M

$I=x_1^{\frac{1}{3}}a^{\frac{2}{3}}$

M=$y_1^{\frac{1}{3}}b^{\frac{2}{3}}$

hence

$\frac{x_1}{a}^{\frac{2}{3}}+\frac{y_1}{b}^{\frac{2}{3}}=1$