First slide

Tangents and normals

Question

$The two curves x^{t} - 3 {xy}^{2} + 2 = 0 and 3 x^{2} y - y^{3} = 2$

Moderate

A

$cuts at right angle$
B

$touch each other$
C

$cut at an angle \frac{π}{3}$
D

$cut at an angle \frac{π}{4}$

Solution

$Two curves cuts at right angle if product of their slopes$ $is - 1 .$
$Two given curves are$
$x^{3} - 3 x y^{2} + 2 = 0 - - - - - - - - - - (1)$
$and 3 x^{2} y - y^{3} - 2 = 0 - - - - - - - - - (2)$
$Differentiate equ. (i),$
$\Rightarrow m_{1} = \frac{d y}{d x} = \frac{3 (x^{2} - y^{2})}{6 x y}$
$Differentiate equ. (ii),$
$\Rightarrow m_{2} = \frac{dy}{dx} = \frac{- 2 xy}{(x^{2} - y^{2})}$
$∴ m_{1} \times m_{2} = \frac{(x^{2} - y^{2})}{2 xy} \times \frac{- 2 xy}{(x^{2} - y^{2})}$
$\Rightarrow m_{1} \times m_{2} = - 1$

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