Question

If the tangent at any point of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ makes an angle $\alpha$ with the major axis and an angle $\beta$ with the focal radius of the point of contact then show that the eccentricity 'e' of the ellipse is given by the absolute value of $\frac{\cos \beta }{\cos \alpha }$.

Answer 1

$p(a\cos \theta ,b\sin \theta )$

equation of tangent $\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$

tangent meet major axis at $T(a\sec \theta ,0)$

Apply sine rule in $\Delta PST$

$\frac{ps}{\left(\sin \right(\pi -\alpha \left)\right)}=\frac{ST}{\sin \beta }$

$\Rightarrow \frac{a(1-e\cos \theta )}{\sin \alpha }=\frac{a\sec \theta -ae}{\sin \beta }\Rightarrow \frac{a(\sec \theta -e)}{\sin \beta }$

$\Rightarrow \frac{a(1-e\cos \theta )}{\sin \alpha }=\frac{a(1-e\cos \theta )}{\cos \theta \sin \beta }$

$\cos \theta =\frac{\sin \alpha }{\sin \beta }\Rightarrow \frac{\sin \beta }{\sin \alpha }=\sec \theta$ __(1)

Slope of tangent

$\frac{-b}{a}\cot \theta =\tan \alpha$

$\frac{-b}{a}=\tan \alpha \tan \theta$

$\frac{b^2}{a^2}={\tan }^2\alpha ({\sec }^2\theta -1)$ now from (1)

$={\tan }^2\alpha \left(\frac{{\sin }^2\beta -{\sin }^2\alpha }{{\sin }^2\alpha }\right)$

$\Rightarrow \frac{b^2}{a^2}=\frac{{\sin }^2\beta -{\sin }^2\alpha }{{\cos }^2\alpha }$ & $e=\sqrt{\frac{1-b^2}{a^2}}$

So $e=\sqrt{1-\frac{{\sin }^2\beta -{\sin }^2\alpha }{{\cos }^2\alpha }}=\sqrt{\frac{1-{\sin }^2\beta }{{\cos }^2\alpha }}=\frac{\cos \beta }{\cos \alpha }=e$