Question

The equation of the hyperbola whose foci are (6,4) and (-4,4) and eccentricity 2. is

• A. $\frac{(x-1{)}^2}{25/4}-\frac{(y-4{)}^2}{75/4}=1$
• B. $\frac{(x+1{)}^2}{25/4}-\frac{(y+4{)}^2}{75/4}=1$
• C. $\frac{(x-1{)}^2}{75/4}-\frac{(y-4{)}^2}{25/4}=1$
• D. none of these

Answer 1

The correct option is A

$\frac{(x-1{)}^2}{25/4}-\frac{(y-4{)}^2}{75/4}=1$

$\text{The centre of the hyperbola is the midpoint of the line joining the two foci}$

$\text{So,the coordinates of the centre are Let 2a and 2b be the length of the transverse and the conjugate axes,respectively.Also,let e be the eccentricity.}$

$\Rightarrow \frac{(x-1{)}^2}{a^2}-\frac{(y-4{)}^2}{b^2}=1$

$Now,distancebetweenthetwofoci=2ae$

$2ae=\sqrt{(6+4{)}^2+(4-4{)}^2}$

$\Rightarrow 2ae=10$

$\Rightarrow ae=5$

$\Rightarrow a=\frac{5}{2}$

$Also,b^2=(ae{)}^2-(a{)}^2$

$\Rightarrow b^2=25-\left(\frac{25}{4}\right)$

$\Rightarrow b^2=\frac{75}{4}$

$\text{Equation of the hyperbola is given below:}$

$\frac{(x-1{)}^2}{25/4}-\frac{(y-4{)}^2}{75/4}=1$