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. - 1/9
• B. - 2/9
• C. - 1/3
• D. - 4/9

Answer 1

Answer: (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) $
$ \Rightarrow 2y - 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} $