Question:
Find the point on the curve $\frac{\mathrm{x}^{2}}{9}+\frac{\mathrm{y}^{2}}{16}=1$ at which the tangents are parallel to $\mathrm{y}$ - axis
Solution:
Since the tangent is parallel to $\mathrm{y}$-axis, its slope is not defined, then the normal is parallel to $\mathrm{x}$-axis whose The Slope is zero.
i.e., $\frac{-1}{\frac{d y}{d x}}=0$
$\Rightarrow \frac{-1}{\frac{-16 x}{9 y}}=0$
$\Rightarrow \frac{-9 y}{16 x}=0$
$\Rightarrow y=0$
Substituting $y=0$ in $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$
$\Rightarrow \frac{x^{2}}{9}+\frac{0^{2}}{16}=1$
$\Rightarrow x^{2}=9$
$\Rightarrow x=\pm 3$
Thus, the required point is $(3,0) \&(-3,0)$