Question

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

Answer 1

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

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

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

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

Parallel to $x-$axis

Given tangent is parallel to $x-$axis

$\Rightarrow$ Slope of tangent$=$Slope of $x-$axis

$\Rightarrow \frac{dy}{dx}=0$

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

This is only possible if $x=0$ (If $y=0,\frac{-16x}{9\times 0}=\infty$)

When $x=0$

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

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

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

$\Rightarrow y^2=16$

$\therefore y=\pm 4$

Hence the points are $(0,4)$ and $(0,-4)$