Question

The angle of intersection to the curve $y = x^2, 6y = 7 - x^3$ at (1, 1) is :

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{3}$
• D. $\pi$

Answer 1

Answer: (A)

$\frac{\pi}{2}$

Answer 2

Let $m_1$ and $m_2$ be slope of curve $y = x^2$ and $6y = 7 - x^3$ respectively. Now, $y = x^2$ $\Rightarrow \frac{dy}{dx} = 2x$ $\Rightarrow \left(\frac{dy}{dx}\right)_{\left(1,1\right)} = 2$ i.e. $m_{1} = 2$ and $6y=7-x^{3}$ $\Rightarrow 6 \frac{dy}{dx} = -3x^{2}$ $\Rightarrow \frac{dy}{dx} = - \frac{3}{6} x^{2} = - \frac{1}{2} x^{2}$ $\Rightarrow \left(\frac{dy}{dx}\right)_{1,1} = - \frac{1}{2}\left(1\right)^{2} = - \frac{1}{2}$ $\therefore m_{2} = - \frac{1}{2}$ $\therefore m_{1} m_{2} = 2 . - \frac{1}{2} = -1$ $\therefore$ Angle of intersection is $90^{\circ}$ i.e.$\frac{\pi}{2}$