If the points $\left( a\cos \alpha ,b\sin \alpha \right)$ and $\left( a\cos \beta ,b\sin \beta \right)$ are end points of a focal chord then find the eccentricity of ellipse in terms of $\alpha $ and $\beta $

Hint: Now we know that the equation of chord joining two point $\left( a\cos \alpha ,b\sin \alpha \right)$ and $\left( a\cos \beta ,b\sin \beta \right)$ is given by $\dfrac{x}{a}\cos \left( \dfrac{\alpha +\beta }{2} \right)+\dfrac{y}{b}\sin \left( \dfrac{\alpha +\beta }{2} \right)=\cos \left( \dfrac{\alpha -\beta }{2} \right)$ . Now we are given that the chord if a focal chord and hence it passes through focus. The coordinate of focus in ellipse is given by $\left( ae,0 \right)$ . Hence, substituting x = ae and y = 0 in the equation of chord we get the value of eccentricity.

Complete step-by-step solution:

Now consider the point $\left( a\cos \alpha ,b\sin \alpha \right)$ and $\left( a\cos \beta ,b\sin \beta \right)$
Let us say $\left( a\cos \alpha ,b\sin \alpha \right)$ is point P and $\left( a\cos \beta ,b\sin \beta \right)$ is point Q.
Now we know that PQ is a focal chord of the ellipse.
Now we need to find the eccentricity of the ellipse.
An eccentricity is a non-negative number that determines the nature of the conic.
Now for the ellipse, the value of eccentricity is always less than 1.
Now the in general the equation of chord joining two point $\left( a\cos \theta ,b\sin \theta \right)$ and $\left( a\cos \phi ,b\sin \phi \right)$ on ellipse is given by $\dfrac{x}{a}\cos \left( \dfrac{\theta +\phi }{2} \right)+\dfrac{y}{b}\sin \left( \dfrac{\theta +\phi }{2} \right)=\cos \left( \dfrac{\theta -\phi }{2} \right)$
Hence we get the equation of chord joining two point $\left( a\cos \alpha ,b\sin \alpha \right)$ and $\left( a\cos \beta ,b\sin \beta \right)$ is given by $\dfrac{x}{a}\cos \left( \dfrac{\alpha +\beta }{2} \right)+\dfrac{y}{b}\sin \left( \dfrac{\alpha +\beta }{2} \right)=\cos \left( \dfrac{\alpha -\beta }{2} \right)..................\left( 1 \right)$
Now we have the equation of chord PQ.
It is given that PQ is a focal chord. A focal chord is nothing but the chord which passes through the focus.
Now we know the focus of ellipse is given by $\left( \pm ae,0 \right)$
Let us say it passes through $\left( ae,0 \right)$
Now substituting x = ae and y = 0 in equation (1) we get
$\dfrac{\left( ae \right)}{a}\cos \left( \dfrac{\alpha +\beta }{2} \right)+\dfrac{\left( 0 \right)}{b}\sin \left( \dfrac{\alpha +\beta }{2} \right)=\cos \left( \dfrac{\alpha -\beta }{2} \right)$
$\Rightarrow e\cos \left( \dfrac{\alpha +\beta }{2} \right)=\cos \left( \dfrac{\alpha -\beta }{2} \right)$
$\therefore e=\dfrac{\cos \left( \dfrac{\alpha -\beta }{2} \right)}{\cos \left( \dfrac{\alpha +\beta }{2} \right)}$
Hence we get the eccentricity of ellipse is $\dfrac{\cos \left( \dfrac{\alpha -\beta }{2} \right)}{\cos \left( \dfrac{\alpha +\beta }{2} \right)}$ .

Note: Note that the equation of chord can also be obtained by using the equation of line by two-point form. The equation of line passing through $\left( {{x}_{1}},{{x}_{2}} \right)$ and $\left( {{y}_{1}},{{y}_{2}} \right)$ is given by $\dfrac{x-{{x}_{1}}}{{{x}_{1}}-{{x}_{2}}}=\dfrac{y-{{y}_{1}}}{{{y}_{1}}-{{y}_{2}}}$ .