Question

Find the angle between the parabola y2=4ax at their point of intersection other than origin.

Answer 1

$=>y^2=4ax..............\left(1\right)$

$=>x^2=4by.................\left(2\right)$

Point of intersection other than $(0,0)$

$=>x^2=4b\left(2\sqrt{ax}\right)$

$=>x^3=(8{)}^2{b}^{2}a$

$=>x=4(a{b}^2{)}^{\frac{1}{3}}=4a^{\frac{1}{3}}.b^{\frac{2}{3}}$

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

Point is $P({4a}^{\frac{1}{3}}.b^{\frac{2}{3}},4a^{\frac{2}{3}}.b^{\frac{1}{3}})$

Tangent at $P$ on $\left(1\right)=>4a^{\frac{2}{3}}.b^{\frac{1}{3}}y=2a(x+4a^{\frac{1}{3}}.b^{\frac{2}{3}})$

Slope, $m_1=\frac{1}{2}{a}^{\frac{1}{3}}.b^{\frac{-1}{3}}$

Tangent at $P$ on $\left(2\right)={4a}^{\frac{1}{3}}.b^{\frac{2}{3}}x=2b(y+{4a}^{\frac{2}{3}}.b^{\frac{1}{3}})$

Slope, $m_2=\frac{1}{2}{b}^{\frac{1}{3}}.a^{\frac{-1}{3}}$

Angle between curves, ie. $\alpha ={\tan }^{-1}\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$={\tan }^{-1}\left|\frac{\frac{1}{2}(\frac{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}-\frac{b^{\frac{1}{3}}}{a^{\frac{1}{3}}})}{1+\frac{1}{4}(\frac{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}.\frac{b^{\frac{1}{3}}}{a^{\frac{1}{3}}})}\right|={\tan }^{-1}\frac{2}{5}\left(\frac{|a^{\frac{2}{3}}-b^{\frac{2}{3}}|}{a^{\frac{1}{3}}.b^{\frac{1}{3}}}\right).$