# Find the point on the curve

Question:

Find the point 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:

Given:

The curve $x^{2}+y^{2}=13$ and the line $2 x+3 y=7$

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

Differentiating the above w.r.t $x$

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

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

$\Rightarrow 2\left(x+y \frac{d y}{d x}\right)=0$

$\Rightarrow\left(x+y \frac{d y}{d x}\right)=0$

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

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

Since, line is $2 x+3 y=7$

$\Rightarrow 3 y=-2 x+7$

$\Rightarrow y=\frac{-2 x+7}{3}$

$\Rightarrow y=\frac{-2 x}{3}+\frac{7}{3}$

$\therefore$ The equation of a straight line is $y=m x+c$, where $m$ is the The Slope of the line.

Thus, the The Slope of the line is $\frac{-2}{3} \ldots(2)$

Since, tangent is parallel to the line,

$\therefore$ the The Slope of the tangent $=$ The Slope of the normal

$\frac{-x}{y}=\frac{-2}{3}$

$\Rightarrow-x=\frac{-2 y}{3}$

$\Rightarrow x=\frac{2 y}{3}$

Substituting $x=\frac{2 y}{3}$ in $x^{2}+y^{2}=13$

$\Rightarrow\left(\frac{2 y}{3}\right)^{2}+y^{2}=13$

$\Rightarrow\left(\frac{4 y^{2}}{9}\right)+y^{2}=13$

$\Rightarrow y^{2}\left(\frac{4}{9}+1\right)=13$

$\Rightarrow y^{2}\left(\frac{13}{9}\right)=13$

$\Rightarrow y^{2}\left(\frac{1}{9}\right)=1$

$\Rightarrow y^{2}=9$

$\Rightarrow y=\pm 3$

Substituting $y=\pm 3$ in $x=\frac{2 y}{3}$, we get,

$x=\frac{2 \times(\pm 3)}{3}$

$x=\pm 2$

Thus, the required point is $(2,3) \&(-2,-3)$