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

$\begin{array}{1 1} (0,4) ; (0,-4) \\ (0,3 ) ; (0,-3) \\ (0,2); (0,-2) \\ (0,1);(0,-1) \end{array}$

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}$$=\large\frac{-16}{9}.\frac{x}{y} Step 2: If the tangents are parallel to x-axis,then \large\frac{dy}{dx}$$=0$
$\Rightarrow \large\frac{-16}{9}.\frac{x}{y}$$=0 Therefore x=0 Step 3: Putting x=0 in equ(1) 0+\large\frac{y^2}{16}$$=1$
$\Rightarrow y=\pm 4$
Therefore the tangents are parallel to $x$-axis at $(0,4)$ and $(0,-4)$.