Question

find the equation of the locus of a point p such that PA²+PB²=2c²,where A=(a,0),B=(-a,0) and 0<|a|<|c|​

Answer 1

Answer:

Step-by-step explanation:

Given,

A(a,0)

B(-a,0)

Let P(x,y)

=PA²+PB²=2C²

Since square and root gets cancelled,

 (a-x)+(0-y)+(-a-x)+(0-y)=2C²

a²+x²-2ax+y²+a²+x²+2ax+y²=2C²

-2ax and +2ax gets cancelled,

2x²+2y²+2a²=2C²

2[x²+y²+a²]=2C²

x²+y²+a²-c²=0