Question

The angle of intersection of $x=\sqrt{y}$ and $x^3+6y=7$ at $(1,1)$ is

• A. $\frac{\pi }{5}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is D $\frac{\pi }{2}$

Let $c_1:x=\sqrt{y}$ and $c_2:x^3+6y=7$
Differentiating w.r.t $x$
$c_1:1=\frac{1}{2\sqrt{y}}\frac{dy}{dx}$ and $c_2:3x^2+6\frac{dy}{dx}=0$
Hence slope of tangent at $(1,1)$ is
$m_1=2$ and $m_2=-\frac{1}{2}$
Clearly $m_1.m_2=-1$
Hence both the curve intersect at right angle.