The eccentric angles of the extremities of a chord of an ellipse x2/a2+y2/b2=1 are . If this chord passes thru the focus then

e=

This is the ans. Can u show the solution with explanation.

Dear Sohini,

Make sure that you use maximum visualisation while solving problems of coordinate. Mastering coordinate geometry is easier comparatively.

It is given that , the eccentric angles of the extermities of a chord of anellipse are ø , and ø₂ , i.e, (a cos ø , b sin ø,)

and (a cos ø₂ , b sin ø_{2 ).}

The equation of this chord is:

y - b sin ø₁ =  (b sin ø₂  - b sin ø₁/a cos ø₂ - a cos ø₁) (x-a cos ø₁)

If this chord is passing through focus, then it should satify the coordinates (ac,0).

Satisfying (ac,0), we get,

0 - b sin ø₁ =  (b sin ø₂  - b sin ø₁/a cos ø₂ - a cos ø₁) (ac-a cos ø₁)

From this we get the eccentricity

e = sin ø₁ + sin ø₂/sin (ø₁ + ø₂ )

Hence, proved