Question

Find the angle of intersection of the following curve:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $x^2+y^2=ab$

Answer 1

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

or $\frac{y^2}{b^2}=1-\frac{x^2}{a^2}$

or $y^2=b^2(1-\frac{x^2}{a^2})$ ........$\left(1\right)$

$y^2=ab-x^2$ ..........$\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$ we get

$ab-x^2=b^2(1-\frac{x^2}{a^2})$

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

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

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

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

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

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

$\Rightarrow x=a\sqrt{\frac{b}{a+b}}$

Put $x=a\sqrt{\frac{b}{a+b}}$ in $\left(2\right)$ we get

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

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

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

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

$\Rightarrow y=b\sqrt{\frac{a}{a+b}}$

Differentiating $\left(1\right)$ w.r.t $x$ we get

$2y\frac{dy}{dx}=-\frac{2x{b}^2}{a^2}$

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

$\Rightarrow {\left[\frac{dy}{dx}\right]}_{(a\sqrt{\frac{b}{a+b}},b\sqrt{\frac{a}{a+b}})}=-\frac{2a\sqrt{\frac{b}{a+b}}\times b^2}{2b\sqrt{\frac{a}{a+b}}{a}^2}$

$m_1=-\frac{a{b}^2\sqrt{b}}{b{a}^2\sqrt{a}}=\frac{b\sqrt{b}}{a\sqrt{a}}$

Differentiating $\left(2\right)$ w.r.t $x$ we get

$2y\frac{dy}{dx}=-2x$

$\Rightarrow \frac{dy}{dx}=\frac{-x}{y}$

$\Rightarrow {\left[\frac{dy}{dx}\right]}_{(a\sqrt{\frac{b}{a+b}},b\sqrt{\frac{a}{a+b}})}=\frac{-a\sqrt{\frac{b}{a+b}}}{b\sqrt{\frac{a}{a+b}}}$

$\Rightarrow m_2=\frac{-a\sqrt{b}}{b\sqrt{a}}$

The angle between the two curves is given as

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

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

$=\left|\frac{(b^2+a^2)\sqrt{ab}}{a^{2}b-b^{2}a}\right|$

$=\left|\frac{(a^2+b^2)\sqrt{ab}}{ab(a-b)}\right|$

$\theta ={\tan }^{-1}\left(\frac{(a^2+b^2)\sqrt{ab}}{ab(a-b)}\right)$