# Find the point on the curve

Question:

Find the point on the curve $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$ at which the tangents are parallel to the

Solution:

$x$ - axis

Given:

The curve is $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$

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

$\Rightarrow \frac{2 x^{2-1}}{4}+\frac{2 y^{2-1} \times \frac{d y}{d x}}{25}=0$

Cross multiplying we get,

$\Rightarrow \frac{25 \times 2 \mathrm{x}+4 \times 2 \mathrm{y} \times \frac{\mathrm{dy}}{\mathrm{ds}}}{100}=0$

$\Rightarrow 50 \mathrm{x}+8 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=0$

$\Rightarrow 8 y \frac{d y}{d x}=-50 x$

$\Rightarrow \frac{d y}{d x}=\frac{-50 x}{8 y}$

$\Rightarrow \frac{d y}{d x}=\frac{-25 x}{4 y} \ldots(1)$

(i)

Since, the tangent is parallel to $x$-axis

$\Rightarrow \frac{d y}{d x}=\tan (0)=0$   .....(2)

$\therefore \tan (0)=0$

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

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

$\Rightarrow \frac{-25 x}{4 y}=0$

$\Rightarrow-25 x=0$

$\Rightarrow x=0$

Substituting $x=0$ in $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$,

$\Rightarrow \frac{0^{2}}{4}+\frac{y^{2}}{25}=1$

$\Rightarrow y^{2}=25$

$\Rightarrow y=\pm 5$

Thus, the required point is $(0,5) \&(0,-5)$

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