# Find the point on the curve

Question:

Find the point on the curve $x^{2}+y^{2}-2 x-3=0$ at which the tangents are parallel to the $x$ - axis

Solution:

Given:

The curve is $x^{2}+y^{2}-2 x-3=0$

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

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

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

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

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

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

(i) Since, the tangent is parallel to $x$-axis

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

$\therefore \tan (0)=0$

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

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

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

$\Rightarrow 1-x=0$

$\Rightarrow x=1$

Substituting $x=1$ in $x^{2}+y^{2}-2 x-3=0$

$\Rightarrow 1^{2}+y^{2}-2 \times 1-3=0$

$\Rightarrow 1+y^{2}-2-3=0$

$\Rightarrow y^{2}-4=0$

$\Rightarrow y^{2}=4$

$\Rightarrow y=\pm 2$

Thus, the required point is $(1,2) \&(1,-2)$