Question

If the parabolas $y^2=4x$ and $x^2=32y$ intersect at $(16,8)$ at an angle $\theta$ , then $\theta$ equals to

• A. $ta{n}^{-1}5/3$
• B. $ta{n}^{-1}4/5$
• C. $ta{n}^{-1}3/5$
• D. $\pi /2$

Answer 1

Answer: (C)

Given curves are
$y^2=4x$
$\Rightarrow 2y\frac{dy}{dx}=4$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(16,8\right)}=\frac{4}{16}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(16,8\right)}\frac{1}{4}=m_1$ (say)
and $x^2=32y$
$\Rightarrow 2x=32\frac{dy}{dx}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(16,8\right)}=\frac{2\times 16}{32}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(16,8\right)}=1=m_2$(say)
$\therefore$ Angle between them, $\theta =ta{n}^{-1}\left(\frac{m_2-m_1}{1+m_1{m}_2}\right)$
$=ta{n}^{-1}\left(\frac{1-\frac{1}{4}}{1+1\times \frac{1}{4}}\right)$
$=ta{n}^{-1}\left(\frac{\frac{3}{4}}{\frac{5}{4}}\right)$
$=ta{n}^{-1}\left(\frac{3}{5}\right)$