Question

Prove that the two parabolas $y^2=4ax$ and $x^2=4by$ intersect (other than the origin) at an angle of ${\tan }^{-1}\left[\frac{3a^{\frac{1}{3}}{b}^{\frac{1}{3}}}{2(a^{\frac{2}{3}}+b^{\frac{2}{3}})}\right]$

Answer 1

Let us first find the points of intersection

$y^2=4ax$ and $x^2=4by$

Therefore

$y=\frac{x^2}{4b}$. Hence ${\left(\frac{x^2}{4b}\right)}^2=4ax$

$\frac{x^4}{16b^2}-4ax=0$

$x^4-64b^{2}ax=0$

$x(x^3-64b^{2}a)=0$

$x=0$ and $x=4b^{\frac{2}{3}}.a^{\frac{1}{3}}$

Therefore, $y=0$ and $y=4b^{\frac{1}{3}}.a^{\frac{2}{3}}$

Hence the point of intersection other than that of origin is $(x,y)=(4b^{\frac{2}{3}}{a}^{\frac{1}{3}},4b^{\frac{1}{3}}{a}^{\frac{2}{3}})$

Now the slope of the tangent at the above point is

$2y{y}^{\prime }=4a$

$y^{\prime }=\frac{4a}{2y}$

$y_{(4b^{\frac{2}{3}}{a}^{\frac{1}{3}},4b^{\frac{1}{3}}{a}^{\frac{2}{3}})}^{\prime }=\frac{4a}{8b^{\frac{1}{3}}{a}^{\frac{2}{3}}}$

$=\frac{b^{\frac{-1}{3}}{a}^{\frac{1}{3}}}{2}$

Similarly

$2x=4b{y}^{\prime }$ or

$y^{\prime }=\frac{x}{2b}$

$y_{(4b^{\frac{2}{3}}{a}^{\frac{1}{3}},4b^{\frac{1}{3}}{a}^{\frac{2}{3}})}^{\prime }=\frac{4b^{\frac{2}{3}}{a}^{\frac{1}{3}}}{2b}=2b^{\frac{-1}{3}}{a}^{\frac{1}{3}}$

Hence the angle between the curves at the point of intersection

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

$=\frac{2b^{\frac{-1}{3}}{a}^{\frac{1}{3}}-\frac{b^{\frac{-1}{3}}{a}^{\frac{1}{3}}}{2}}{1+\frac{b^{\frac{-1}{3}}{a}^{\frac{1}{3}}}{2}.2b^{\frac{-1}{3}}{a}^{\frac{1}{3}}}$

$=\frac{3b^{\frac{-1}{3}}{a}^{\frac{1}{3}}}{2(1+b^{\frac{-2}{3}}{a}^{\frac{2}{3}})}$

$=\frac{3b^{\frac{1}{3}}{a}^{\frac{1}{3}}}{2(b^{\frac{2}{3}}+a^{\frac{2}{3}})}$

Or

$\theta ={\tan }^{-1}\left[\frac{3b^{\frac{1}{3}}{a}^{\frac{1}{3}}}{2(b^{\frac{2}{3}}+a^{\frac{2}{3}})}\right]$