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

The given equation is,

$y^2+x^2-2x-3=0$

which can be rewritten as,

$y^2=-x^2+2x+\mathrm{3...}\left(i\right)$

Slope of the tangent is given by $\frac{dy}{dx}$.

Finding the derivative, differentiate both sides with respect to x.

Therefore,

$2y\frac{dy}{dx}=-2x+2\frac{dy}{dx}=\frac{1-x}{y}$

For a tangent parallel to the x-axis.

Slope=0.

That is,

$\frac{dy}{dx}=\frac{1-x}{y}=01-x=0x=1$

Put this value of x in (i) to get the corresponding values of y.

Hence,

$y^2=-1^2+2\cdot 1+3y^2=4y=\pm 2$

Therefore the required points are,

$P(1,2)Q(1,-2)$