Question

Given the points $A(0,4)$ and $B(0,-4)$, the equation of the locus of the point $P$ such that $|AP-BP|=6$ is-

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

Answer 1

Answer: (A)

$9x^2-7y^2+63=0$

We have,

$|AP-BP|=6$

$\Rightarrow AP-BP=\pm 6$

$\Rightarrow AP=BP\pm 6$

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

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

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

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

$\Rightarrow {(4y+9)}^2=9(x^2+y^2+8y+16)$

$\Rightarrow 16y^2+72y+81=9(x^2+y^2+8y+16)$

$\Rightarrow 7y^2-9x^2-63=0$

$\Rightarrow 9x^2-7x^2+63=0$

Hence, this is the answer.