Question

At what point, the slope of the tangent to the curve

$x^{2} + y^{2} - 2x - 3 = 0$is zero :

• A. (3, 0); (-1, 0)
• B. (3,0) ( 1,2)
• C. (-1, 0) (1,2)
• D. (1,2) (1, -2)

Answer 1

Answer: (D)

(1,2) (1, -2)

Answer 2

Given $x^{2} + y^{2} - 2x - 3 = 0$ ………….(1)

Diff. w.r.t. x, we get

$2x + 2y\frac{dy}{dx} - 2 = 0$

⇒ $\frac{dy}{dx} = \frac{1 - x}{y}$

Since the slope of the tangent to the curve is zero.

$\therefore$ $\frac{dy}{dx} = 0$

⇒ $1 - x = 0$

⇒ x = 1

Put $x = 1$ in equation (1), we get $1 + y^{2} - 2 - 3 = 0$

⇒ $y^{2} = 4$

⇒ $y = \pm \ 2$

Hence regd. points are (1,2) and (1, -2)