Question

Find the angle between two diameters of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ Whose extremities have eccentricity angle a and $\beta =a+\frac{\pi }{2}$

Answer 1

Let ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$SlopeofOP$ $=m_1=\frac{b\sin a}{a\cos a}=\frac{b}{a}\tan a$

$SlopeofOQ=m_1=\frac{b\sin \beta }{a\cos \beta }=-\frac{b}{a}\cot a$

$(given\beta =a+\frac{\pi }{2})$

$\therefore \tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{\frac{b}{a}(\tan a+\cot a)}{1-\frac{b^2}{a^2}}\right|=\left|\frac{2ab}{(a^2-b^2\sin 2a)}\right|$