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

This is second part of the multi-part question q13

Toolbox:
• If $y=f(x)$,then $\big(\large\frac{dy}{dx}\big)_P$=slope of the tangent to $y=f(x)$ at the point $P$.
• Equation of line with given two points is $\large\frac{x-x_1}{x_2-x_1}=\large\frac{y-y_1}{y_2-y_1}$
Step 1:
Given :$\large\frac{x^2}{9}+\frac{y^2}{16}$$=1-----(1) Now differentiating w.r.t x we get, \large\frac{2x}{9}+\frac{2y}{16}\frac{dy}{dx}$$=0$
$\Rightarrow \large\frac{dy}{dx}$$=\large\frac{-16}{9}.\frac{x}{y} Step 2: If the tangents are parallel to y-axis,then \large\frac{dx}{dy}$$=0$
Therefore $y=0$
Step 3:
Putting $y=0$ in equ(1)