Question

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

Answer 1

 The centre of the hyperbola is the midpoint of the line joining  the two focii.
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 conjugate axis and let e be the eccentricity.

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

Distance between the two focii = 2ae

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

\[ \Rightarrow 2ae = 10\]

\[ \Rightarrow ae = 5\]

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

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:

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

\[ \Rightarrow \frac{4\left( x^2 - 2x + 1 \right)}{25} - \frac{\left( y^2 - 8y + 16 \right)}{75} = 1\]

\[ \Rightarrow \frac{\left( 4 x^2 - 8x + 4 \right)}{25} - \frac{\left( 4 y^2 - 32y + 64 \right)}{75} = 1\]

\[ \Rightarrow 3\left( 4 x^2 - 8x + 4 \right) - \left( 4 y^2 - 32y + 64 \right) = 75\]

\[ \Rightarrow 12 x^2 - 24x + 12 - 4 y^2 + 32y - 64 = 75\]

\[ \Rightarrow 12 x^2 - 24x - 4 y^2 + 32y - 127 = 0\]