Question

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

Answer 1

Given curve: $\frac{x^2}{9}+\frac{y^2}{16}=1\cdots \left(i\right)$
Differentiating w.r.t. $x$,
we get $\frac{2x}{9}+\frac{2y}{16}\times \frac{dy}{dx}=0$
$\Rightarrow \frac{x}{9}+\frac{y}{16}\times \frac{dy}{dx}=0$
$\Rightarrow \frac{y}{16}\times \frac{dy}{dx}=-\frac{x}{9}$
$\Rightarrow \frac{dy}{dx}=-\frac{16}{9}\left(\frac{x}{y}\right)\cdots \left(ii\right)$
Tangent is parallel to x-axis, then slope of the line is $0$,
$\Rightarrow \frac{dy}{dx}=0$
$\Rightarrow -\frac{16}{9}\left(\frac{x}{y}\right)=0$
$\Rightarrow x=0$
From equation (i), we get
$\frac{x^2}{9}+\frac{y^2}{16}=1$
$\Rightarrow \frac{y^2}{16}=1$
$\Rightarrow y^2=16$
$\Rightarrow y=\pm 4$
Hence, the points are $(0,4)$ and $(0,-4)$
Tangent is parallel to y-axis,
then normal is parallel to x-axis,
so slope of normal is zero.
Slope of normal $=-\frac{1}{\text{Slope of tangent}}=-\frac{dx}{dy}=0$
$\Rightarrow \frac{9}{16}\left(\frac{y}{x}\right)=0$
$\Rightarrow y=0$
From equation (i), we get
$\frac{x^2}{9}+\frac{y^2}{16}=1$
$\Rightarrow \frac{x^2}{9}=1$
$\Rightarrow x^2=9$
$\Rightarrow x=\pm 3$
Hence, the points are $(3,0)$ and $(-3,0)$.
So, the required points are $(0,\pm 4)$ and $(\pm ,3,0)$.