Question

What is the Equation of chord joining point ${\theta }_1$ & ${\theta }_2$ an the Eclipse. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Answer 1

Point $P$ is $\left(a\cos {\theta }_1,bsin{\theta }_1\right)$

and point $Q$ is $\left(a\cos {\theta }_2,b\sin {\theta }_2\right)$

Equation of the chord is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

$\Rightarrow y-bsin{\theta }_1=\frac{b\sin {\theta }_2-bsin{\theta }_1}{a\cos {\theta }_2-\cos {\theta }_1}\left(x-a\cos {\theta }_1\right)$

$\Rightarrow$$y-bsin{\theta }_1=\frac{b}{a}\left(\frac{\sin {\theta }_2-sin{\theta }_1}{\cos {\theta }_2-\cos {\theta }_1}\right)\left(x-a\cos {\theta }_1\right)$

$\begin{array}{c}\Rightarrow b\sin \theta -b\sin {\theta }_1=\frac{b}{a}\left(\frac{\sin {\theta }_2-\sin {\theta }_1}{\cos {\theta }_2-\cos {\theta }_1}\right)\times \left(acos\theta -a\cos {\theta }_1\right)\\ \Rightarrow \frac{\left(b\sin \theta -b\sin {\theta }_1\right)}{\left(acos\theta -a\cos {\theta }_1\right)}=\frac{b}{a}\left(\frac{\sin {\theta }_2-\sin {\theta }_1}{\cos {\theta }_2-\cos {\theta }_1}\right)\\ \Rightarrow \frac{\sin \theta -\sin {\theta }_1}{cos\theta -\cos {\theta }_1}=\frac{\sin {\theta }_2-\sin {\theta }_1}{\cos {\theta }_2-\cos {\theta }_1}\end{array}$