Find the points on the curve

Solution:

(i) The slope of the $x$-axis is 0 .

Now, let $\left(x_{1}, y_{1}\right)$ be the required point.

Since, the point lies on the curve.

Hence, $\frac{x_{1}{ }^{2}}{9}+\frac{y_{1}{ }^{2}}{16}=1 \quad \ldots(1)$

Now,

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

$\Rightarrow \frac{2 x}{9}+\frac{2 y}{16} \frac{d y}{d x}=0$

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

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

Now,

Slope of the tangent at $(x, y)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{-16 x_{1}}{9 y_{1}}$

Slope of the tangent at $(x, y)=$ Slope of the $x$-axis [Given]

$\therefore \frac{-16 x_{1}}{9 y_{1}}=0$

$\Rightarrow x_{1}=0$

$0+\frac{y_{1}{ }^{2}}{16}=1 \quad[$ From eq. (1) $]$

Also,

$y_{1}^{2}=16$

$y_{1}=\pm 4$

Thus, the required points are $(0,4)$ and $(0,-4)$.

(ii) The slope of the $y$-axis is $\infty$.

Let $\left(x_{1}, y_{1}\right)$ be the required point.

Given:

Since, the point lies on the curve.

Hence, $\frac{x_{1}{ }^{2}}{9}+\frac{y_{1}{ }^{2}}{16}=1$           ....(1)

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

$\Rightarrow \frac{2 x}{9}+\frac{2 y}{16} \frac{d y}{d x}=0$

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

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

Now,

Slope of the tangent at $(x, y)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{-16 x_{1}}{9 y_{1}}$

Slope of the tangent at $\left(x_{1}, y_{1}\right)=$ Slope of the $y$-axis [Given]

$\therefore \frac{-16 x_{1}}{9 y_{1}}=\infty$

$\Rightarrow \frac{9 y_{1}}{-16 x_{1}}=0$

$\Rightarrow y_{1}=0$

$\Rightarrow \frac{x_{1}^{2}}{9}+0=1$  $[$ From eq. (1)]

$\Rightarrow x_{1}^{2}=9$

$\Rightarrow x_{1}=\pm 3$

Thus, the required points are $(3,0)$ and $(-3,0)$.