Question

Equation of the locus of all points such that the difference of its distances from $(-3,-7)$ and $(-3,3)$ is $8$ units is

• A. $\frac{(x+3{)}^2}{16}-\frac{(y+2{)}^2}{9}=1$
• B. $\frac{(x+3{)}^2}{9}-\frac{(y+2{)}^2}{16}=-1$
• C. $\frac{(x+3{)}^2}{9}-\frac{(y+2{)}^2}{19}=1$
• D. $\frac{(x+3{)}^2}{7}-\frac{(y+2{)}^2}{19}=-1$

Answer 1

The correct option is B $\frac{(x+3{)}^2}{9}-\frac{(y+2{)}^2}{16}=-1$

Let $(-3,-7)$ and $(-3,3)$ be $A$ and $B$, respectively and $P$ be any point on hyperbola.
Given, $|PA-PB|=8$ unit
Thus $|PA-PB|=8<10\text{units}\left(\text{Distance between}A\text{and}B\right)$
So, locus of point $P$ is Hyperbola having foci at $A,B$ and transverse axis parallel to $y$ axis as abscissa of foci are same
Here $|PA-PB|=2b=8(\text{where}b=\text{length of semi transverse axis})$
$\Rightarrow b=4$
Distance between foci $=2be=10$
$\Rightarrow e=\frac{5}{b}=\frac{5}{4}$
From $e=\sqrt{1+\frac{a^2}{b^2}},(\text{where}a=\text{semi conjugate axis length})$
$a=\sqrt{b^2(e^2-1)}=3$

Therefore equation of the locus is $\frac{(x+3{)}^2}{9}-\frac{(y+2{)}^2}{16}=-1$