 # Find the points on the curve `
Question:

Find the points on the curve $x^{2}+y^{2}=13$, the tangent at each one of which is parallel to the line $2 x+3 y=7$.

Solution:

Let $\left(x_{1}, y_{1}\right)$ represent the required point.

The slope of line $2 x+3 y=7$ is $\frac{-2}{3}$.

Since, the point lies on the curve.

Hence, $x_{1}^{2}+y_{1}^{2}=13 \quad \cdots$ (1)

Now, $x^{2}+y^{2}=13$

On differentiating both sides w.r.t. $x$, we get

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

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

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

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

$\Rightarrow \frac{-x_{1}}{y_{1}}=\frac{-2}{3}$

$\Rightarrow x_{1}=\frac{2 y_{1}}{3}$    .....(2)

From eq. (1), we get

$\left(\frac{2 y_{1}}{3}\right)^{2}+y_{1}^{2}=13$

$\Rightarrow \frac{13 y_{1}{ }^{2}}{9}=13$

$\Rightarrow y_{1}{ }^{2}=9$

$\Rightarrow y_{1}=\pm 3$

$\Rightarrow y_{1}=3$ or $y_{1}=-3$

and

$x_{1}=2$ or $x_{1}=-2$ [From eq. (2)]

Thus, the required points are $(2,3)$ and $(-2,-3)$