Question

Find points on the curve $\frac{x^2}{9}+\frac{y^2}{16}=1$ at which the tangents are

(a) parallel to X-axis

(b) parallel to Y-axis.

Answer 1

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

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

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

(a) For tangent parallel to X-axis, we must have, $\frac{dy}{dx}=0$

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

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

$\frac{0^2}{9}+\frac{y^2}{16}=1\Rightarrow y^2=16\Rightarrow y=\pm 4$

Hence, the points on Eq. (i) at which the tangents are parallel to X-axis are (0,4) and (0,-4)

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)

(b) 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).