Question

If ${\left(\frac{a}{b}\right)}^{1/3}+{\left(\frac{b}{a}\right)}^{1/3}=\frac{\sqrt{3}}{2}$, then the angle of intersection of the parabola $y^2=4ax$ and $x^2=4by$ at the point other than the origin is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is B $\frac{\pi }{3}$

The points of intersection of the two parabola are $(0,0)$ and $(4\sqrt[3]{3a{b}^2},4\sqrt[3]{3a^{2}b})$

Let $m_1$ and $m_2$ be the slope of tangents to parabola $\Rightarrow y^2=4ax$ and $x^2=4by$ respectively.

$\Rightarrow m_1=\frac{dy}{dx}=\frac{2a}{y}$

$\Rightarrow m_2=\frac{dy}{dx}=\frac{x}{2b}$

So, the slope of tangents to the parabola at $(4\sqrt[3]{3a{b}^2},4\sqrt[3]{3a^{2}b})$ is

$\Rightarrow m_1=\frac{2a}{y}=\frac{2a}{4\sqrt[3]{3a^{2}b}}=\frac{1}{2}\sqrt[3]{3\frac{a}{b}}$

$\Rightarrow m_2=\frac{x}{2b}=\frac{4\sqrt[3]{3a{b}^2}}{2b}=2\sqrt[3]{3\frac{a}{b}}$

Let $\theta$ be the angle between the two tangents

$\therefore tan\theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$=\left|\frac{\frac{1}{2}\sqrt[3]{3\frac{a}{b}}-2\sqrt[3]{3\frac{a}{b}}}{1+\left(\frac{1}{2}\sqrt[3]{3\frac{a}{b}}\right)\left(2\sqrt[3]{3\frac{a}{b}}\right)}\right|$

$=\left|\frac{-\frac{3}{2}\sqrt[3]{3\frac{a}{b}}}{1+\left(\sqrt[3]{3\frac{a}{b}}\right)\left(\sqrt[3]{3\frac{a}{b}}\right)}\right|$

$=\left|\frac{-\frac{3}{2}}{\sqrt[3]{3\frac{b}{a}}+\sqrt[3]{3\frac{a}{b}}}\right|$

$=\left|\frac{-\frac{3}{2}}{\frac{\sqrt{3}}{2}}\right|$

$=\sqrt{3}$

$\therefore \theta =ta{n}^{-1}\sqrt{3}=\frac{\pi }{3}$

So, the answer is option (B)