 Click to Chat

1800-1023-196

+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
`        find the locus of  a point P such that the sum of it's distances from (0,2) and (0,-2)  is 6.`
4 months ago

```							 Let point be P(h,k)((h – 0)^2 + (k – 2)^2)^1/2 + ((h – 0)^2 + (k + 2)^2)^1/2 = 6(h^2 + (k – 2)^2)^1/2 + (h^2 + (k + 2)^2)^1/2 = 6(h^2 + (k – 2)^2)^1/2  = 6 – (h^2 + (k + 2)^2)^1/2now square both sidesh^2 + k^2 – 4k + 4  = 36 + h^2 + k^2 + 4k + 4 – 12(h^2 + (k + 2)^2)^1/2– 4k  = 36 + 4k – 12(h^2 + (k + 2)^2)^1/28k + 36 = 12(h^2 + (k + 2)^2)^1/22k + 9 = 3((h^2 + (k + 2)^2)^1/2)square both sides4k^2 + 81 + 36k = 9(h^2 + k^2 + 4 + 4k)9h^2 + 5k^2 = 459x^2 + 5y^2 = 45 Regards
```
4 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Analytical Geometry

View all Questions »  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 53 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions