# Find the point on the curve

Question:

Find the point on the curve $\frac{\mathrm{x}^{2}}{9}+\frac{\mathrm{y}^{2}}{16}=1$ at which the tangents are parallel to $\mathrm{x}$ - axis

Solution:

Given:

The curve is $\frac{\mathrm{x}^{2}}{9}+\frac{\mathrm{y}^{2}}{16}=1$

Differentiating the above w.r.t $x$, we get the Slope of tangent,

$\Rightarrow \frac{2 x^{2-1}}{9}+\frac{2 y^{2-1} \times \frac{d y}{d s}}{16}=0$

$\Rightarrow \frac{2 x}{9}+\frac{y \times \frac{d y}{d x}}{8}=0$

Cross multiplying we get,

$\Rightarrow \frac{(8 \times 2 x)+(9 \times y) \times \frac{d y}{d x}=0}{72}$

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

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

$\Rightarrow \frac{d y}{d x}=\frac{-16 x}{9 y} \ldots(1)$

(i)

Since, the tangent is parallel to $\mathrm{x}$ - axis]

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\tan (0)=0 \ldots(2)$

$\therefore \tan (0)=0$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=$ The Slope of the tangent $=\tan \theta$

From (1) \& (2), we get,

$\Rightarrow \frac{-16 x}{9 y}=0$

$\Rightarrow-16 x=0$

$\Rightarrow x=0$

Substituting $x=0$ in $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$

$\Rightarrow \frac{0^{2}}{9}+\frac{y^{2}}{16}=1$

$\Rightarrow y^{2}=16$

$\Rightarrow y=\pm 4$

Thus, the required point is $(0,4) \&(0,-4)$