Question

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

• A. $\frac{x^2}{7}-\frac{y^2}{9}=1$
• B. $\frac{y^2}{9}-\frac{x^2}{7}=1$
• C. $\frac{x^2}{9}-\frac{y^2}{7}=1$
• D. $\frac{y^2}{7}-\frac{x^2}{9}=1$

Answer 1

The correct option is B $\frac{y^2}{9}-\frac{x^2}{7}=1$

$|AP-BP|=6$

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

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

Squaring it, we get

$[x^2+(y{)}^2-8y+16]=[x^2+(y{)}^2+8y+16]+36\pm 12\sqrt{x^2+(y+4{)}^2}$

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

$\Rightarrow 4y+9=\pm 3\sqrt{x^2+(y+4{)}^2}$

Squaring this, we get

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

$\Rightarrow \frac{y^2}{9}-\frac{x^2}{7}=1$.