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

#### Options

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

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

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

none of these

#### Solution

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

The centre of the hyperbola is the midpoint of the line joining the two foci.

So, the coordinates of the centre are \[\left( \frac{6 - 4}{2}, \frac{4 + 4}{2} \right), i . e . \left( 1, 4 \right) .\]

Let 2a and 2b be the length of the transverse and the conjugate axes, respectively. Also, let e be the eccentricity.

\[\Rightarrow \frac{\left( x - 1 \right)^2}{a^2} - \frac{\left( y - 4 \right)^2}{b^2} = 1\]

Now, distance between the two foci = 2*ae*

\[2ae = \sqrt{\left( 6 + 4 \right)^2 + \left( 4 - 4 \right)^2}\]

\[ \Rightarrow 2ae = 10\]

\[ \Rightarrow ae = 5\]

\[ \Rightarrow a = \frac{5}{2}\]

\[\text { Also }, b^2 = \left( ae \right)^2 - \left( a \right)^2 \]

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

\[ \Rightarrow b^2 = \frac{75}{4}\]

Equation of the hyperbola is given below:

\[\frac{\left( x - 1 \right)^2}{25/4} - \frac{\left( y - 4 \right)^2}{75/4} = 1\]