Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Shohini Sinha Ray Grade: Upto college level
```
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.
```
8 years ago

8 Points
```										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

```
8 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Analytical Geometry

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details