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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 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 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
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
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