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 $y$ - axis.

Solution:

Since, the tangent is parallel to $y$ - axis, its The Slope is not defined, then the normal is parallel to $x$-axis whose The Slope is zero.

i.e, $\frac{-1}{\frac{d y}{d x}}=0$

$\Rightarrow \frac{-1}{\frac{-25 x}{4 y}}=0$

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

$\Rightarrow y=0$

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

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

$\Rightarrow x^{2}=4$

$\Rightarrow x=\pm 2$

Thus, the required point is $(2,0) \&(-2,0)$

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