Question

The equation of the hyperbola whose foci are at $(4,6)$, $(4,-4)$ respectively and having eccentricity $2$ is

• A. $\frac{(y-1{)}^2}{25/8}-\frac{(x-4{)}^2}{75/4}=1$
• B. $\frac{(y-1{)}^2}{25/4}-\frac{(x-4{)}^2}{75/4}=1$
• C. $\frac{(y-1{)}^2}{75/4}-\frac{(x-4{)}^2}{25/4}=1$
• D. $\frac{(y-1{)}^2}{75/4}-\frac{(x-4{)}^2}{25/8}=1$
  

Answer 1

The correct option is B $\frac{(y-1{)}^2}{25/4}-\frac{(x-4{)}^2}{75/4}=1$

The centre of the hyperbola is the midpoint of the line joining the two foci. So the coordinates of the centre are $(\frac{4+4}{2},\frac{6-4}{2})$ i.e. $(4,1)$
Since here, the $x$ coordinate for foci is same, therefore transverse axis lies parallel to $y-$ axis.
The equation of hyperbola with centre at $(4,1)$ is $\frac{(y-1{)}^2}{a^2}-\frac{(x-4{)}^2}{b^2}=1$
$\because$ Distance between the foci $=2ae$
$\sqrt{(4-4{)}^2+(6+4{)}^2}=2a\times 2$
$\Rightarrow 10=4a\Rightarrow a=\frac{5}{2}$
$\therefore b^2=a^2(e^2-1)=\frac{25}{4}(4-1)=\frac{75}{4}$
Thus the equation of the hyperbola is $\frac{(y-1{)}^2}{25/4}-\frac{(x-4{)}^2}{75/4}=1$