Question

The angle between the curves $y^2 = 4ax$ and $ay = 2x^2$ is

• A. $\tan^{-1} \frac{3}{4}$
• B. $\tan^{-1} \frac{3}{5}$
• C. $\tan^{-1} \frac{4}{3}$
• D. $\tan^{-1} \frac{5}{3}$

Answer 1

Answer: (B)

$\tan^{-1} \frac{3}{5}$

Answer 2

We have, $y^2 = 4ax$ .....(i)
and, $ay = 2x^2$ ....(ii)
Solving (i) and (ii), we get
$x = a$ and $y = 2a$
Differentiating (i) w.r.t $'x'$, we get
$2y \frac{dy}{dx} = 4a$
Now, $\left[ \frac{dy}{dx} \right]_{(a,2a)} = \frac{4a}{2(2a)} = 1$
$\Rightarrow \:\:\: m_1 = 1$
Now, differentiating (ii) w.r.t 'x', we get
$a \frac{dy}{dx} = 4x$
$\Rightarrow \:\:\: \left[ \frac{dy}{dx} \right]_{(a,2a)} = \frac{4a}{a} = 1$
$\Rightarrow \:\: m_2 = 4$
$\because$ Angle between curves is equal to angle between their tangents.
$\Rightarrow \:\: \tan \theta =\left|\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}\right| \Rightarrow \tan \theta =\left|\frac{4-1}{1+4 \times\left(1\right)}\right|$
$\Rightarrow \:\: \tan \theta = \frac{3}{5} \Rightarrow \:\: \theta = \tan^{-1} \frac{3}{5}$