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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


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

7 years ago


Answers : (1)


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 ø2 ).

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




7 years ago

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