# Find the angle to intersection of the following curves:

Question:

Find the angle to intersection of the following curves:

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and $x^{2}+y^{2}=a b$

Solution:

Given:

Curves $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ .......(1)

$\& x^{2}+y^{2}=a b \ldots(2)$

Second curve is $x^{2}+y^{2}=a b$

$y^{2}=a b-x^{2}$

Substituting this in equation (1),

$\Rightarrow \frac{x^{2}}{a^{2}}+\frac{a b-x^{2}}{b^{2}}=1$

$\Rightarrow \frac{x^{2} b^{2}+a^{2}\left(a b-x^{2}\right)}{a^{2} b^{2}}=1$

$\Rightarrow x^{2} b^{2}+a^{3} b-a^{2} x^{2}=a^{2} b^{2}$

$\Rightarrow x^{2} b^{2}-a^{2} x^{2}=a^{2} b^{2}-a^{3} b$

$\Rightarrow x^{2}\left(b^{2}-a^{2}\right)=a^{2} b(b-a)$

$\Rightarrow x^{2}=\frac{a^{2} b(b-a)}{x^{2}\left(b^{2}-a^{2}\right)}$

$\Rightarrow x^{2}=\frac{a^{2} b(b-a)}{x^{2}(b-a)(b+a)}$

$\Rightarrow x^{2}=\frac{a^{2} b}{(b+a)}$

$\therefore a^{2}-b^{2}=(a+b)(a-b)$

$\Rightarrow X=\pm \sqrt{\frac{a^{2} b}{(b+a)}} \ldots(3)$

since, $y^{2}=a b-x^{2}$

$\Rightarrow y^{2}=a b-\left(\frac{a^{2} b}{(b+a)}\right)$

$\Rightarrow y^{2}=\frac{a b^{2}+a^{2} b-a^{2} b}{(b+a)}$

$\Rightarrow y^{2}=\frac{a b^{2}}{(b+a)}$

$\Rightarrow y=\pm \sqrt{\frac{a b^{2}}{(b+a)}} \ldots(4)$

since, curves are $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \& x^{2}+y^{2}=a b$

Differentiating above w.r.t $\mathrm{x}$,

$\Rightarrow \frac{2 x}{a^{2}}+\frac{2 y}{b^{2}} \cdot \frac{d y}{d x}=0$

$\Rightarrow \frac{y}{b^{2}} \cdot \frac{d y}{d x}=-\frac{x}{a^{2}}$

$\Rightarrow \frac{d y}{d x}=\frac{\frac{x}{a^{2}}}{\frac{y^{2}}{b^{2}}}$

$\Rightarrow \frac{d y}{d x}=\frac{-b^{2} x}{a^{2} y}$

$\Rightarrow m_{1}=\frac{d y}{d x}=\frac{-b^{2} x}{a^{2} y} \ldots(5)$

Second curve is $x^{2}+y^{2}=a b$

$\Rightarrow 2 x+2 y \cdot \frac{d y}{d x}=0$

$\Rightarrow m_{2}=\frac{d y}{d x}=\frac{-x}{y} \ldots(6)$

Substituting (3) in (4), above values for $m_{1} \& m_{2}$, we get,

At $\left(\sqrt{\frac{\mathrm{a}^{2} \mathrm{~b}}{(\mathrm{~b}+\mathrm{a})}}, \sqrt{\frac{\mathrm{ab}^{2}}{(\mathrm{~b}+\mathrm{a})}}\right)$ in equation $(5)$, we get

$\Rightarrow \frac{d y}{d x}=\frac{-b^{2} \times \sqrt{\frac{a^{2} b}{(b+a)}}}{a^{2} \times \sqrt{\frac{a b^{2}}{(b+a)}}}$

$\Rightarrow \frac{d y}{d x}=\frac{-b^{2} \times a \sqrt{\frac{b}{(b+a)}}}{a^{2} \times b \sqrt{\frac{a}{(b+a)}}}$

$\Rightarrow \frac{d y}{d x}=\frac{-b^{2} a \sqrt{b}}{a^{2} b \sqrt{a}}$

$\Rightarrow m_{1}=\frac{d y}{d x}=\frac{-b \sqrt{b}}{a \sqrt{a}}$

At $\left(\sqrt{\frac{a^{2} b}{(b+a)}}, \sqrt{\frac{a b^{2}}{(b+a)}}\right)$ in equation $(6)$, we get

$\Rightarrow \frac{d y}{d x}=\frac{-\sqrt{\frac{a^{2} b}{(b+a)}}}{\sqrt{\frac{a b^{2}}{(b+a)}}}$

$\Rightarrow \frac{d y}{d x}=\frac{-a \sqrt{\frac{b}{(b+a)}}}{b \sqrt{\frac{a}{(b+a)}}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{a} \sqrt{\mathrm{b}}}{\mathrm{b} \sqrt{\mathrm{a}}}$

$\Rightarrow m_{2}=\frac{d y}{d x}=-\sqrt{\frac{a}{b}}$

when $m_{1}=\frac{-b \sqrt{b}}{a \sqrt{a}} \& m_{2}=-\sqrt{\frac{a}{b}}$

$\Rightarrow \tan \theta=\left|\frac{\frac{-b \sqrt{b}}{a \sqrt{a}}-\sqrt{\frac{a}{b}}}{1+\frac{-b \sqrt{b}}{a \sqrt{a}} \times-\sqrt{\frac{a}{b}}}\right|$

$\Rightarrow \tan \theta=\left|\frac{\frac{-b \sqrt{b}}{a \sqrt{a}}+\sqrt{\frac{a}{b}}}{1+\frac{b}{a}}\right|$

$\Rightarrow \tan \theta=\left|\frac{\frac{-b \sqrt{b} \times \sqrt{b}+a \sqrt{a} \times \sqrt{a}}{a \sqrt{2} \times \sqrt{b}}}{1+\frac{b}{a}}\right|$

$\Rightarrow \tan \theta=\left|\frac{\frac{-b \times b+a \times a}{a \sqrt{2} b}}{1+\frac{b}{a}}\right|$

$\Rightarrow \tan \theta=\left|\frac{\frac{\mathrm{a}^{2}-\mathrm{b}^{2}}{\mathrm{a} \sqrt{\mathrm{a}} \mathrm{b}}}{\mathrm{a}+\mathrm{b}}\right|$

$\Rightarrow \tan \theta=\left|\frac{\frac{(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})}{\sqrt{\mathrm{a}} \mathrm{b}}}{\mathrm{a}+\mathrm{b}}\right|$

$\Rightarrow \tan \theta=\left|\frac{(\mathrm{a}-\mathrm{b})}{\sqrt{\mathrm{a}} \mathrm{b}}\right|$

$\Rightarrow \theta=\tan ^{-1}\left(\frac{(a-b)}{\sqrt{a b}}\right)$