Question

Consider the parametric equation $x=\frac{a(1-t^2)}{1+t^2},y=\frac{2at}{1+t^2}$.

What is $\frac{d^{2}y}{d{x}^2}$ equal to?

• A. $\frac{a^2}{y^2}$
• B. $\frac{a^2}{x^2}$
• C. $-\frac{a^2}{x^2}$
• D. $-\frac{a^2}{y^3}$

Answer 1

The correct option is D $-\frac{a^2}{y^3}$

$x=\frac{a(1-t^2)}{1+t^2},y=\frac{2at}{1+t^2}$

Let $t=\tan \theta$

$\Rightarrow x=a\sin 2\theta ,y=a\cos 2\theta \Rightarrow x^2+y^2=a^2$

Differentiating the above equation gives

$2x+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{-x}{y}$

Differentiating again wrt $x$ gives

$\frac{d^{2}y}{{dx}^2}=-\frac{y-x\left(\frac{dy}{dx}\right)}{y^2}=\frac{-(x^2+y^2)}{y^3}=\frac{-a^2}{y^3}$