Question

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

Answer 1

$\frac{x^2}{9}+\frac{y^2}{16}=1$

Differentiating both sides w.r.t $x$ we get

$\frac{2x}{9}+\frac{2y}{16}\frac{dy}{dx}=0$

$\Rightarrow \frac{2y}{16}\frac{dy}{dx}=\frac{-2x}{9}$

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

Given:The line is parallel to $y-$axis

$\Rightarrow$ Angle with $x-$axis$={90}^{\circ }$

$\Rightarrow \theta ={90}^{\circ }$

Slope$=\tan \theta =\tan {90}^{\circ }=\infty$

Hence $\frac{dy}{dx}=\infty$

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

$\Rightarrow \frac{16x}{9y}=\frac{1}{0}$

This will be possible only if the denominator is $0$

$\Rightarrow 9y=0$

$\Rightarrow y=0$

Put $y=0$ in $\frac{x^2}{9}+\frac{y^2}{16}=1$ we get

$\frac{x^2}{9}=1$

$\Rightarrow x^2=9$

$\Rightarrow x=\pm 3$

Hence the points at which tangent is parallel to $y-$axis are $(3,0)$ and $(-3,0)$