Question

Find points on the curve $\frac{x^2}{9}+\frac{y^2}{16}=1$ =1 at which the tangents are parallel to Y-axis.

Answer 1

The equation of the given curve is $\frac{x^2}{9}+\frac{y^2}{16}=1$ =1 ..(i)

On differentiating both sides w.r.t. x, we get

$\frac{2x}{9}+\frac{1}{16}\left(2y\frac{dy}{dx}\right)=0\Rightarrow \frac{y}{8}\frac{dy}{dx}=-\frac{2x}{9}\Rightarrow \frac{dy}{dx}=-\frac{16x}{9y}$ ...(ii)

For tangents parallel to Y-axis, we must have, $\frac{dx}{dy}=0$

$\Rightarrow -\frac{9y}{16x}=0\Rightarrow y=0$

When y=0, then from Eq. (i), we get

$\frac{x^2}{9}+\frac{0^2}{16}=1\Rightarrow x^2=9\Rightarrow x=\pm 3$

Hence, the points on Eq. (i) at which the tangents are parallel to X-axis are (3,0) and (-3,0).