Question

Find points on the curve ${\displaystyle \frac{x^2}{9}+\frac{{\text{y}}^2}{16}=1}$ at which the tangent is parallel to x-axis.

Answer 1

The equation of the given curve is ${\displaystyle \frac{x^2}{9}+\frac{{\text{y}}^2}{6}=1}$

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

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

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

The tangent is parallel to the x-axis if the slope of the tangent is i.e., ${\displaystyle 0 \frac{-16x}{\text{9y}}=0,}$ which is possible if x = 0.

Then, ${\displaystyle \frac{x^2}{9}+\frac{{\text{y}}^2}{6}=1}$ for x = 0

⇒ y² = 16

⇒ y = ± 4

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