Question

Find points on the curve $\frac{x^2}{9}+\frac{y^2}{16}=1$ at which the tangents are
(i) Parallel to $x$-axis .
(ii) 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 with respect to $x$, we have:
$\frac{2x}{9}+\frac{2y}{16}\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{-16x}{9y}$

(i)
The tangent is parallel to the $x$-axis if the slope of the tangent is $0.$
i.e.,$\frac{-16x}{9y}=0$, which is possible if $x=0.$.
Then, $\frac{x^2}{9}+\frac{y^2}{16}=1$
for $x=0,$ $\Rightarrow =y^2=16\Rightarrow y=\pm 4$
Hence, the points at which the tangents are parallel to the $x$-axis are $(0,4)$ and $(0,-4)$.

(ii)

The tangent is parallel to the $y$-axis if the slope of the normal is $0$,
which gives $\frac{-1}{\left(\frac{-16}{9y}\right)}=\frac{9y}{16x}=0$ $\Rightarrow y=0$
Then, $\frac{x^2}{9}+\frac{y^2}{16}=1$ for $y=0$, $\Rightarrow x=\pm 3$
Hence, the points at which the tangents are parallel to the $y$-axis are $(3,0)$ and $(-3,0)$.