Question

Find the points on the curve $\frac{x^2}{16}+\frac{y^2}{49}=1$ at which the tangents are
(i) parallel to x-axis
(ii) parallel to y-axis.

Answer 1

Given the curve $\frac{x^2}{16}+\frac{y^2}{49}=1$

$\frac{y^2}{49}=1-\frac{x^2}{16}\frac{d\left(\frac{y^2}{49}\right)}{dx}=\frac{d(1-\frac{x^2}{16})}{dx}\frac{2ydy}{49dx}=\frac{-2x}{16}\frac{dy}{dx}=\frac{-49x}{16y}$

$\left(i\right)$ Parallel to x-axis: slope of tangent is equal to slope of x-axis

$\frac{dy}{dx}=0$

$\frac{-49x}{16y}=0$

$x=0$

Therefore substituting x value in the equation of the curve we get

$\frac{0}{16}+\frac{y^2}{49}=1$

$y^2=49y=\pm 7$

Hence the points are $(0,-7)$ and $(0,7)$

$\left(ii\right)$ Parallel to y-axis: slope of tangent is equal to slope of y-axis

$\frac{dy}{dx}=\infty$

$\frac{-49x}{16y}=\infty$

$y=0$

Substituting y value in curve we get

$\frac{x^2}{16}+\frac{0}{49}=1x^2=16x=\pm 4$

Therefore the points are $(0,-4)$ and $(0,4)$