Question

If x = a sec θ, y = b tan θ, prove that $\frac{d^{2}y}{d{x}^2}=-\frac{b^4}{a^2{y}^3}$.

Answer 1

Here,
$$\begin{gathered} x=a\sec \theta \mathrm{and}y=b\tan \theta \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\theta ,\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}x}{\mathit{d}\theta }=a\sec \theta \tan \theta \mathrm{and}\frac{\mathit{d}y}{\mathit{d}\theta }=b{\sec }^2\theta \\ \therefore \frac{dy}{dx}=\frac{dy}{d\theta }\times \frac{d\theta }{dx}=\frac{b{\sec }^2\theta }{a\sec \theta \tan \theta }=\frac{b\mathrm{cosec}\theta }{a} \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\frac{b}{a}\times -\mathrm{cosec}\theta \cot \theta \times \frac{d\theta }{dx} \\ =-\frac{b}{a}\times \mathrm{cosec}\theta \cot \theta \times \frac{1}{a\sec \theta \tan \theta } \\ =-\frac{b}{a^2}\times \cot \theta \times \frac{1}{{\tan }^2\theta } \\ =-\frac{b}{a^2}\times \frac{1}{{\tan }^3\theta } \\ =\frac{-b^4}{a^2{y}^3}\left[\because y=b\tan \theta \right] \end{gathered}$$

Hence proved.