Question

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

• A. $\theta ={\tan }^{-1}\frac{3}{8}$
• B. $\theta ={\tan }^{-1}\frac{5}{8}$
• C. $\theta ={\tan }^{-1}\frac{3}{7}$
• D. $\theta ={\tan }^{-1}1$

Answer 1

The correct option is A $\theta ={\tan }^{-1}\frac{3}{8}$

Given parabolas,
$y^2=4ax$ and $x^2=4by$
$\Rightarrow \frac{y^4}{16a^2}=4by\Rightarrow y=4a^{2/3}{b}^{1/3}$

Intersection point of the given parabolas is $(0,0)$ and $(4a^{1/3}{b}^{2/3},4a^{2/3}{b}^{1/3})$
Slope of the tangent on the parabola $y^2=4ax$ at $(4a^{1/3}{b}^{2/3},4a^{2/3}{b}^{1/3})$ is
$m_1=\frac{2a}{y}=\frac{2a}{4a^{2/3}{b}^{1/3}}=\frac{a^{1/3}}{2b^{1/3}}$

Slope of the tangent on the parabola $x^2=4by$ at $(4a^{1/3}{b}^{2/3},4a^{2/3}{b}^{1/3})$ is
$m_2=\frac{x}{2b}=\frac{4a^{1/3}{b}^{2/3}}{2b}=\frac{2a^{1/3}}{b^{1/3}}$

So, the angle between the parabola is
$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{\frac{a^{\frac{1}{3}}}{2b^{\frac{1}{3}}}-\frac{2a^{\frac{1}{3}}}{b^{\frac{1}{3}}}}{1+\frac{a^{\frac{2}{3}}}{b^{\frac{2}{3}}}}\right|=\frac{3}{2}\times \frac{1}{{\left(\frac{a}{b}\right)}^{\frac{1}{3}}+{\left(\frac{b}{a}\right)}^{\frac{1}{3}}}\Rightarrow \tan \theta =\frac{3}{2}\times \frac{1}{4}\Rightarrow \theta ={\tan }^{-1}\frac{3}{8}$