Question

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

Parallel to $x$-axis are $(a,\pm b)$.Find $a+b$

Answer 1

The equation of the given curve is $\frac{x^2}{9}+\frac{y^2}{16}=1$

On differentiating both sides with respect to $x$, we have:

$\frac{2x}{9}+\frac{2y}{16}\cdot \frac{dy}{dx}=0$

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

The tangent is parallel to the $x$-axis if the slope of the tangent is $0$

i.e.,$\frac{-16x}{9y}=0$, which is possible if $x=0$.

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

for $x=0,$ $\Rightarrow =y^2=16\Rightarrow y=\pm 4$

Hence, the points at which the tangents are parallel to the $x$-axis are $(0,4)$ and $(0,4)$.