# Find the points on the curve

Question:

Find the points on the curve $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$ at which the tangents are parallel to the (i) $x$-axis (ii) $y$-axis.

Solution:

(i) The slope of the $x$-axis is 0 .

Now, let $\left(x_{1}, y_{1}\right)$ be the required point.

Since, the point lies on the curve.

Hence, $\frac{x_{1}{ }^{2}}{4}+\frac{y_{1}{ }^{2}}{25}=1 \quad \ldots(1)$

Now, $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$

$\therefore \frac{2 x}{4}+\frac{2 y}{25} \frac{d y}{d x}=0$

$\Rightarrow \frac{2 y}{25} \frac{d y}{d x}=\frac{-x}{2}$

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

Now,

Slope of the tangent at $\left(x_{1}, y_{1}\right)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{-25 x_{1}}{4 y_{1}}$

Slope of the tangent at $\left(x_{1}, y_{1}\right)=$ Slope of the $x$-axis [Given]

$\therefore \frac{-25 x_{1}}{4 y_{1}}=0$

$\Rightarrow x_{1}=0$

Also,

$0+\frac{y_{1}{ }^{2}}{25}=1$     $[$ From eq. (1) $]$

$\Rightarrow y_{1}^{2}=25$

$\Rightarrow y_{1}=\pm 5$

Thus, the required points are $(0,5)$ and $(0,-5)$

(ii) The slope of the $y$-axis is $\infty$.

Now, let $\left(x_{1}, y_{1}\right)$ be the required point.

Since, the point lies on the curve.

Hence, $\frac{x_{1}{ }^{2}}{4}+\frac{y_{1}{ }^{2}}{25}=1 \quad \ldots(1)$

Now, $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$

$\therefore \frac{2 x}{4}+\frac{2 y}{25} \frac{d y}{d x}=0$

$\Rightarrow \frac{2 y}{25} \frac{d y}{d x}=\frac{-x}{2}$

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

Now,

Slope of the tangent at $\left(x_{1}, y_{1}\right)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{-25 x_{1}}{4 y_{1}}$

Slope of the tangent at $\left(x_{1}, y_{1}\right)=$ Slope of the $y$-axis [Given]

$\therefore \frac{-25 x_{1}}{4 y_{1}}=\infty$

$\Rightarrow \frac{4 y_{1}}{-25 x_{1}}=0$

$\Rightarrow y_{1}=0$

Also,

$\frac{x_{1}^{2}}{4}=1$       [From eq. (1)]

$\Rightarrow x_{1}^{2}=4$

$\Rightarrow x_{1}=\pm 2$

Thus, the required points are $(2,0)$ and $(-2,0)$.