Question

If the parabola $y^2=4x$ and $x^2=32y$ intersect at $\left(16,8\right)$ at an angle $\theta$

• A. ${\tan }^{-1}\left(\frac{3}{5}\right)$
• B. ${\tan }^{-1}\left(\frac{4}{5}\right)$
• C. $\pi$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (A)

${\tan }^{-1}\left(\frac{3}{5}\right)$

$y^2=4x$ and $x^2=32y$ intersect at (16,8)

at an angle $\theta$

Now slope of the tangent to $y^2=4x$ at (16,8)

is given by $m_1=(\frac{dy}{dx}{)}_{(16,8)}=(\frac{4}{2y}{)}_{(16,8)}$

$=\frac{1}{4}$

Again slope of tangent to $x^2=32y$ at (16,8)

is given by $m_2=(\frac{dy}{dx}{)}_{(16,8)}=(\frac{2x}{32}{)}_{(16,8)}=1.$

We know $tan\theta =\frac{m_2-m_1}{m_2+m_1}$

$=\frac{1-\frac{1}{4}}{1+\frac{1}{4}}=\frac{3}{5}$

$\therefore \theta =ta{n}^{-1}\left(\frac{3}{5}\right)$