Question

Given the points $A(0,4)$ and $B(0,-4)$ then the locus of the point $P(x, y)$ such that $|A P-B P|=4$, 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 $A (0,4) \& B (0,-4)$
$\therefore AB =2 be =8$
$\&| AP - BP |=$ length of transverse $axis =2 b =4 \Rightarrow b =2$
$\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 $