Find points on the curve $\cfrac{{{x^2}}}{9} + \cfrac{{{y^2}}}{{16}} = 1$ at which the tangents are

(i) parallel to $x$-axis

(ii) parallel to $y$-axis.

We have,
$\cfrac{{{x^2}}}{9} + \cfrac{{{y^2}}}{{16}} = 1$ …(1)

Differentiating (1) w.r.t. $x$, we get
$\cfrac{{2x}}{9} + \cfrac{1}{{16}}\left( {2y\cfrac{{dy}}{{dx}}} \right) = 0 \Rightarrow \cfrac{y}{8}\cfrac{{dy}}{{dx}} = - \cfrac{{2x}}{9}$

$\Rightarrow \cfrac{{dy}}{{dx}} = - \cfrac{{16x}}{{9y}}$ …(2)

(i) For tangents parallel to $x$-axis, we must have
$\cfrac{{dy}}{{dx}} = 0 \Rightarrow - \cfrac{{16x}}{{9y}} = 0 = x = 0$

When $x = 0$, then from (1),

$\cfrac{{{0^2}}}{9} + \cfrac{{{y^2}}}{{16}} = 1 \Rightarrow {y^2} = 16 \Rightarrow y = \pm 4$

Therefore the points on (1) at which the tangents are parallel to $x$-axis are $(0,\dot 4)$ and $(0,\; - 4)$ .

(ii) For tangents parallel to $y$-axis, we must have $\cfrac{{dx}}{{dy}} = 0$

$\Rightarrow - \cfrac{{9y}}{{16x}} = 0 \Rightarrow y = 0$

When $y = 0$, then from (1), $\cfrac{{{x^2}}}{9} + \cfrac{{{0^2}}}{{16}} = 1 \Rightarrow {x^2} = 9 \Rightarrow x = \pm 3$

Therefore, The points on (1) at which the tangents are parallel to $y$-axis are$(3,\;0)$ and $( - 3,0)$ .