Question

The angle of intersection between curve $xy = 6$and $x^{2}y = 12$

• A. $\tan^{- 1}\left( \frac{3}{4} \right)$
• B. $\tan^{- 1}\left( \frac{3}{11} \right)$
• C. $\tan^{- 1}\left( \frac{11}{3} \right)$
• D. $0^{o}$

Answer 1

Answer: (B)

$\tan^{- 1}\left( \frac{3}{11} \right)$

Answer 2

The equation of two curves are $xy = 6$ and $x^{2}y = 12$ from (i) we obtain $y = \frac{6}{x}$ putting this value of $y$ in equation (ii) to obtain $x^{2}\left( \frac{6}{x} \right) = 12$ ⇒ $6x = 12$ ⇒ $x = 2$

Putting $x = 2$ in (i) or (ii) we get, $y = 3.$ Thus, the two curves intersect at P(2, 3)

Differentiating (i) w.r.t. x, we get $x\frac{dy}{dx} + y = 0$ ⇒ $\frac{dy}{dx} = \frac{- y}{x}$

⇒ $\left( \frac{dy}{dx} \right)_{(2,3)} = - \frac{3}{2} = m_{1}$

Differentiating (ii) w.r.t. x, we get $x^{2}\frac{dy}{dx} + 2xy = 0$

⇒ $\frac{dy}{dx} = \frac{- 2y}{x}$

⇒ $\left( \frac{dy}{dx} \right)_{(2,3)} = - 3 = m_{2}$

⇒ $\tan\frac{\theta = \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} = \left( \frac{- 3}{2} + 3 \right)}{\left( 1 + \left( \frac{- 3}{2} \right)( - 3) \right)} = \frac{3}{11}$

⇒$\theta = \tan^{- 1}\frac{3}{11}$.