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
```        If P is any point lying on the ellipse x2/a2 +y2/b2=1,a>b with foci S and S',then the locus of the incentre of the triangle PSS' is
a) (1-e)x2+(1+e)y2=a2
b)(1+e)x2+(1-e)y2=a2c2
c)(1+e)x2+(1-e)y2=a2
d)(1-e)x2+(1+e)y2=a2e2(1-e)```
8 years ago

147 Points
```										Dear Vaibhav
x2/a2 +y2/b2=1
S=(ae,o)
S' =(-ae,o)
let a point P =(a cosΘ,bsinΘ)
we now focul distance SP=a-ex=a-eacosΘ
and  S'P = a+ex = a+excosΘ
and SS'=2ae
so incenter of triangle x=(SS' acosΘ + SP (-ae) +S'P(ae))/( SP +S'P+SS')
x=aecosΘ
simply for y co ordinate  y= e(bsinΘ)/(1+e)
so cosΘ =x/ae
and sinΘ =y(1+e)/eb
square and add locus will come (1-e)x2+(1+e)y2=a2e2(1-e)

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.
```
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