Question

If $p$ and $q$ are intercepts on the axes by any tangent line on the curve $\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1$, then show that $\frac{p}{a}+\frac{q}{b}=1$

Answer 1

The given curve is $\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1$

Now, at any point $(h,k)$ on this curve, the slope of the tangent is

$\frac{1}{2\sqrt{ax}}+\frac{1}{2\sqrt{by}}\frac{dy}{dx}=0$

$\therefore \frac{dy}{dx}{|}_{(h,k)}=-\sqrt{\frac{bk}{ah}}$

Hence, equation of tangent through $(h,k)$ is

$y=-\sqrt{\frac{bk}{ah}}x+c$ (where $c$ is a constant)

$\therefore k=-\sqrt{\frac{bk}{ah}}h+c$ (as it passes through $(h,k)$)

$\therefore c=k+\sqrt{\frac{bk}{ah}}h$

Now, x-intercept of tangent is $p$ hence,

$0=-\sqrt{\frac{bk}{ah}}p+k+\sqrt{\frac{bk}{ah}}h$

$\therefore p=\frac{k+\sqrt{\frac{bk}{ah}}h}{\sqrt{\frac{bk}{ah}}}$

Now, y-intercept of tangent is $q$ hence,

$q=k+\sqrt{\frac{bk}{ah}}h$

$⟹\frac{p}{a}+\frac{q}{b}=\frac{k}{a}\sqrt{\frac{ah}{bk}}+\frac{h}{a}+\frac{k}{b}+\sqrt{\frac{hk}{ab}}$

$=\sqrt{\frac{hk}{ab}}+\frac{h}{a}+\frac{k}{b}+\sqrt{\frac{hk}{ab}}$

$=\frac{h}{a}+2\sqrt{\frac{hk}{ab}}+\frac{k}{b}$

$={(\sqrt{\frac{h}{a}}+\sqrt{\frac{k}{b}})}^2$

$=1$ (from putting $(h,k)$ in equation of curve and squaring)

Hence, proved.