Question

If tangent to the curve $y^2 = x^3$ at its point $(m^2, m^3)$ is also normal to the curve at $(M^2, M^3)$, then what is the value of mM ?

• A. -0.111111111
• B. -0.222222222
• C. -0.333333333
• D. -0.444444444

Answer 1

The Correct Option is D

Equation of the given curve is $y^2 = x^3$ ...(1) On differentiating with respect to x $2y \frac{dy}{dx}=3x^{2} $ $ \Rightarrow \frac{dy}{dx} = \frac{3x^{2}}{2y}$ Now , $ \left(\frac{dy}{dx}\right)_{\left(m^{2} , m^{3}\right) } = \frac{3m^{4}}{2m^{3}} = \frac{3}{2}m$ and $ \left(\frac{dy}{dx}\right)_{\left(M^{2} , M^{3}\right)} = \frac{3M^{4}}{2M^{3}} = \frac{3}{2} M $ Equation of tangents at point $ \left(m^{2} , m^{3}\right) $ is $ \left(y -m^{3}\right) = \frac{3}{2} m\left(x - m^{2}\right) $ $ \Rightarrow2y - 2m^{3} = 3 mx - 3m^{3} $ $ \Rightarrow 3 mx - 2y = 3m^{3} - 2m^{3}$ $ \Rightarrow 3 mx - 2y = m^{3} $ ....(2) Equation of normal at point $(M^2, M^3)$ is $ \left(y -M^{3}\right) = - \frac{2}{3M} \left(x - M^{2}\right)$ $ \Rightarrow 3My -3M^{4} = -2x + 2M^{2} $ $ \Rightarrow 2x + 3 My = 3M^{4} + 2M^{2} $ .....(3) Since, equation (2) and (3) are same $ \Rightarrow \frac{3m}{2} = \frac{-2}{3M} = \frac{m^{3} }{3M^{4} + 2M^{2}} $ $ \Rightarrow \frac{3m}{2} = - \frac{2}{3M} $ $ \Rightarrow mM = - \frac{4}{9} $