Pedal form to polar form of an ellipse

Question

How do I convert the equation

$$\frac{a^2b^2}{p^2}=a^2+b^2-\frac{1}{u^2}$$

into the following equivalent form?

$$u^2=\frac{\sin^2 \theta}{b^2}+\frac{\cos^2 \theta}{a^2}$$

where $$\frac{1}{p^2}=u^2+\left( \frac{du}{d\theta} \right)^2$$

EDIT:

I have tried and found out

$$\frac{uab}{\sqrt{u^2(a^2+b^2)-1-u^4a^2b^2}}\, du = d\theta$$

How do I integrate and express the result in terms of $\sin \theta$ and $\cos \theta$?

$p$ is the perpendicular distance from $O$ to the tangent line to $C$ at the point in case of pedal equation of a curve. It is converted to $u$ as stated above.

Answer 1

Assume $a>b>0$,

\begin{align} t &= \sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta} \\ t^2 &= (a^2-b^2)\sin^2 \theta+b^2 \\ \sin^2 \theta &= \frac{t^2-b^2}{a^2-b^2} \\ t^2 &= a^2+(b^2-a^2)\cos^2 \theta \\ \cos^2 \theta &= \frac{a^2-t^2}{a^2-b^2} \\ dt &= \frac{(a^2-b^2)\sin \theta \cos \theta} {\sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta}} \, d\theta \\ &=\frac{\sqrt{(t^2-b^2)(a^2-t^2)}}{t} \, d\theta \\ d\theta &= \frac{t \, dt}{\sqrt{(t^2-b^2)(a^2-t^2)}} \\ \end{align}

Now

\begin{align} d\theta &= \frac{abu}{\sqrt{(a^2 u^2-1)(1-b^2 u^2)}} \, du \\ &= \frac{abu \times (ab\, du)}{\sqrt{(a^2 b^2 u^2-b^2)(a^2-a^2b^2u^2)}} \\ &= \frac{t \, dt}{\sqrt{(t^2-b^2)(a^2-t^2)}} \\ dt &= ab \, du \end{align}

With boundary conditions:

• $\theta=0$ , $u=\dfrac{1}{r}=\dfrac{1}{b}$
• $\theta=\dfrac{\pi}{2}$ , $u=\dfrac{1}{r}=\dfrac{1}{a}$

Therefore,

$$\fbox{$u=\frac{\sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta}}{ab}$}$$

Alternatively, \begin{align} \int \frac{t\, dt}{\sqrt{(a^2-t^2)(t^2-b^2)}} &= \tan^{-1} \sqrt{\frac{t^2-b^2}{a^2-t^2}} \\ \int \frac{u\, du}{\sqrt{(a^2 u^2-1)(1-b^2 u^2)}} &= \frac{1}{ab} \tan^{-1} \left( \frac{b}{a} \sqrt{\frac{a^2 u^2-1}{1-b^2 u^2}} \right) \end{align}