Question

A(4,0), B(-4,0) are two points. The locus of P which moves such that |PA-PB| =6 is

• A. $9x^2-7y^2+63=0$
• B. $9x^2+7y^2-63=0$
• C. $9x^2+7y^2+63=0$
• D. $9x^2-7y^2-63=0$

Answer 1

The correct option is A $9x^2-7y^2+63=0$

Given,

$A(4,0),B(-4,0)$

Let $P(x,y)$

$|PA-PB|=6$

$\sqrt{(x-4{)}^2+y^2}-\sqrt{(x+4{)}^2+y^2}=6$

$\sqrt{(x-4{)}^2+y^2}=6+\sqrt{(x+4{)}^2+y^2}$

squaring on both sides, we get,

$(x-4{)}^2+y^2=36+(x+4{)}^2+y^2+12\sqrt{(x+4{)}^2+y^2}$

$-16x-36=12\sqrt{(x+4{)}^2+y^2}$

$-4(4x+9)=12\sqrt{(x+4{)}^2+y^2}$

$16x^2+81+72x=9\left[\right(x+4{)}^2+y^2]$

$7x^2-9y^2=63$