First slide

Tangents and normals

Question

$If the tangent at any point P (4 m^{2}, 8 m^{3}) of x^{3} - y^{2} = 0 is also a normal to curve$ $x^{3} - y^{2} = 0, then m =$

Moderate

A

$\pm \frac{\sqrt{2}}{3}$
B

$\pm \frac{\sqrt{3}}{2}$
C

$\frac{3}{\sqrt{2}}$
D

$- \frac{3}{\sqrt{2}}$

Solution

$\begin{array}{l} x^{3} - y^{2} = 0 \dots \dots \dots (1) \\ Differentiating with respect to x \end{array}$
$\begin{array}{l} 2 y \frac{d y}{d x} = 3 x^{2} \\ ∴ Slope of tangent at P = \frac{d y}{d x} = {(\frac{3 x^{2}}{2 y})}_{(4 m^{2}, 8 m^{3})} = 3 m \end{array}$
$\begin{array}{l} ∴ Equation of tangent at P is y - 8 m^{3} = 3 m (x - 4 m^{2}) \\ Or y = 3 m x - 4 m^{3} \dots \dots . . (2) \end{array}$
$\begin{array}{l} Solving (1) and (2) \\ (i.e) x^{3} = y^{2} and y = 3 m x - 4 m^{3} \\ {(3 m x - 4 m^{3})}^{2} = x^{3} \\ x^{3} - 9 m^{2} x^{2} + 24 m^{4} x - 16 m^{6} = 0 \end{array}$
By trial and error method, $x = m^{2}$ satisfies above equations.
After further factorization we get
$\begin{array}{l} {(x - 4 m^{2})}^{2} (x - m)^{2} = 0 \\ ∴ x = 4 m^{2}, m^{2} \\ Put x = m^{2} in (2); then y = - m^{3} \\ ∴ Q (m^{2}, - m^{3}) \end{array}$
$\begin{array}{l} Slope of tangent at Q = {(\frac{d y}{d x})}_{(m^{2}, - m^{3})} = - \frac{3}{2} m \\ Slope of normal at Q = \frac{2}{3 m} \end{array}$
$∴ \frac{2}{3 m} = 3 m$
$\begin{array}{l} (Or) 9 m^{2} = 2 \\ ∴ m = \pm \frac{\sqrt{2}}{3} \\ Therefore, the correct answer is (1). \end{array}$

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