# Find points on the curve

Question:

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

(i) parallel to x-axis (ii) parallel to y-axis

Solution:

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:

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

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

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

(ii) The tangent is parallel to the $y$-axis if the slope of the normal is 0, which gives $\frac{-1}{\left(\frac{-16 x}{9 y}\right)}=\frac{9 y}{16 x}=0 \Rightarrow y=0$.

Then, $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$ for $y=0$

$\Rightarrow x=\pm 3$

Hence, the points at which the tangents are parallel to the y-axis are

(3, 0) and (− 3, 0).