Question

The angle of intersection between the curves $y^2=8xand{x}^2=4y-12$ at $(2,4)$ is :

• A. ${90}^{\circ }$
• B. ${60}^{\circ }$
• C. ${45}^{\circ }$
• D. $0^{\circ }$

Answer 1

The correct option is D $0^{\circ }$

Let $c_1:y^2=8x$ and $c_2:x^2=4y-12$
Differentiating w.r.t $x$
$c_1:2y\frac{dy}{dx}=8$ and $c_2:2x=4\frac{dy}{dx}$
${\left(\frac{dy}{dx}\right)}_{c_1}=\frac{4}{y}$ and ${\left(\frac{dy}{dx}\right)}_{c_2}=-\frac{x}{2}$
Thus slope of tangent at $(2,4)$ to the given curve are $m_1=1$ and $m_2=1$
Therefore, angle of intersection between the given curve is, $\theta ={\tan }^{-1}\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=0^0$
Hence, option 'D' is correct.