Question

Find the equation of the hyperbola whose  foci are (4, 2) and (8, 2) 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 \[\left( \frac{4 + 8}{2}, \frac{2 + 2}{2} \right), i . e . \left( 6, 2 \right)\]

Let 2a and 2b be the length of the transverse and conjugate axis. Let e be the eccentricity.

\[\Rightarrow \frac{\left( x - 6 \right)^2}{a^2} - \frac{\left( y - 2 \right)^2}{b^2} = 1\]

Distance between the two focii = 2ae

\[2ae = \sqrt{\left( 4 - 8 \right)^2 + \left( 2 - 2 \right)^2}\]

\[ \Rightarrow 2ae = 4\]

\[ \Rightarrow ae = 2\]

\[ \Rightarrow a = 1\]

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

\[ \Rightarrow b^2 = 4 - 1\]

\[ \Rightarrow b^2 = 3\]

Equation of the hyperbola:

\[\frac{\left( x - 6 \right)^2}{1} - \frac{\left( y - 2 \right)^2}{3} = 1\]

\[ \Rightarrow \frac{\left( x^2 - 12x + 36 \right)}{1} - \frac{\left( y^2 - 4y + 4 \right)}{3} = 1\]

\[ \Rightarrow 3\left( x^2 - 12x + 36 \right) - \left( y^2 - 4y + 4 \right) = 3\]

\[ \Rightarrow 3 x^2 - 36x + 108 - y^2 + 4y - 4 = 3\]

\[ \Rightarrow 3 x^2 - y^2 - 36x + 4y + 101 = 0\]