Question

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

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is D ${90}^{\circ }$

Given,

$x^2=4ay$

$y^2=4ax$

$\Rightarrow {\left(\frac{x^2}{4a}\right)}^2=4ax$

$x^4=64a^{3}x$

$x^4-64a^{3}x=0$

$x(x^3-64a^3)=0$

$x=0,4a$

when, $x=0\to y=0$

when, $x=4a\to y=4a$

Now,

$y^2=4ax$

$2y\frac{dy}{dx}=4a$

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

when, $x=2a,y=2a$

$m_1=\frac{2a}{y}=\frac{2a}{2a}=1$..........(1)

$x^2=4ay$

$2x=4a\frac{dy}{dx}$

$m_2=\frac{dy}{dx}=\frac{x}{2a}$..........(2)

when, $x=2a,y=2a$

$m_2=\frac{x}{2a}=\frac{2a}{2a}=1$..........(1)

Since,

$\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}$

$=\frac{1-1}{1+\left(1\right)\left(1\right)}$

$=\infty$

$\theta ={\tan }^{-1}\infty ={90}^{\circ }$