Question

The $x-$intercept of the tangent at any arbitrary point of the curve $\frac{a}{x^2}+\frac{b}{y^2}=1$ is proportional to

Answer 1

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

Now, differentiating above equation we get

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

Let $(p,q)$ be any arbitrary point of $\tan$ gency on the curve, then equation of tangent at $(p,q)$ is

$y-q=-\frac{a{q}^3}{b{p}^3}(x-p)$

So, the $x-$intercept of the tangent at $(p,q)$ is obtained by putting $y=0$ in the above equation

$i.e,x=\frac{b{p}^3}{a{q}^2}+p$........$\left(1\right)$

Now as $(p,q)$ lies on the given curve so it satisfies the equation $\frac{a}{x^2}+\frac{b}{y^2}=1$

$i.e,a{q}^2+b{p}^2=p^2{q}^2$......$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get

$x=\frac{b{p}^3+pa{q}^2}{a{q}^2}=\frac{p(a{q}^2+b{p}^2)}{a{q}^2}=\frac{p^2{q}^2}{a{q}^2}=\frac{p^3}{a}$

Hence, $x$ is proportional to $p^3$