Question

A(a,0),B(-a,0),C(c,0) are three points. The equation of the locus of a point P such that PA^2 + PB^2 = 2PC^2 is........

Answer 1

Answer:

x=a^2 -c^2/2c

Step-by-step explanation:

A(a,0

B(-a,0)

C(c,0)

let P(x,y)

distance between two points (x1,y1),(x2,y2)=√(x2-x1)^2+(y2-y1)^2

PA=√(a-x)^2+(y)^2

PB=√(-a-x)^2+(y)^2

PC=√(c-x)^2+(y)^2

PA^2 + PB^2 = 2PC^2

(√(a-x)^2+(y)^2)+(√(-a-x)^2+(y)^2)=2(√(c-x)^2+(y)^2)^2

(a-x)^2+(y)^2+(-a-x)^2+(y)^2)=2((c-x)^2+(y)^2)^2

a^2+x^2-2ax+y^2 a^2+x^2+2ax+y^2=2(c^2+x^2-2cx+y^2)

a^2+x^2-2ax+y^2 a^2+x^2+2ax+y^2=2c^2+2x^2-4cx+2y^2

2a^2+2x^2+2y^2 =2c^2+2x^2-4cx+2y^2

2a^2 =2c^2-4cx (taking 2 as common)

a^2 =c^2-2cx

2cx=a^2 -c^2

x=a^2 -c^2/2c