Question

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

• A. $\frac{x^2}{4}+\frac{y^2}{12}=1$
• B. $\frac{x^2}{12}-\frac{y^2}{4}=1$
• C. $\frac{x^2}{4}-\frac{y^2}{12}=1$
• D. $\frac{x^2}{12}-\frac{y^2}{4}=-1$

Answer 1

Answer: (D)

Distance $AB=8$
$\therefore$ Difference of distances of point $P$ from $A\&B$ is constant & less than distance between fixed points
$\therefore$ Locus of $P$ is a hyperbola with foci $A(0,4)\&B(0,-4)$
$\therefore AB=2be=8$
$\&|AP-BP|=$ length of transverse $axis=2b=4\Rightarrow b=2$
$\therefore e=2$
$\because a^2=b^2\left(e^2-1\right)=4(4-1)=12$
$\&$ centre of hyperbola $=(0,0)$
$\therefore \text{ Equation of hyperbola }\Rightarrow \frac{x^2}{12}-\frac{y^2}{4}=-1$