Question

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

Answer 1

Given curve is, $x^2+y^{2}2x-3=0$ ...(1)
Differentiating w.r.t. $x\frac{d(x^2+y^2-2x-3)}{dx}0$
$\Rightarrow 2x+2y\times \frac{dy}{dx}-2=0$
$\Rightarrow 2y\times \frac{dy}{dx}=2-2x$
$\Rightarrow \frac{dy}{dx}=\frac{2-2x}{2y}$
$\Rightarrow \frac{dy}{dx}=\frac{1-x}{y}$ ...(2)
Tangents are parallel to the x-axis, then the slope $=0$
$\frac{dy}{dx}=0$
$\Rightarrow \frac{1-x}{y}=0$
$\Rightarrow x=1$
Putting $x=1$ in equation (1)
$(1{)}^2+y^2-2(1)-3=0$
$\Rightarrow 1+y^2-2-3=0$
$\Rightarrow y^2-4=0$
$\Rightarrow y^2=4$
$\Rightarrow y=\pm 2$
Hence, the required points are $(1,2)\&(1,-2)$.