Question

Find the equation of the hyperbola whose vertices are (−8, −1) and (16, −1) and focus is (17, −1).

Answer 1

The centre of the hyperbola is given below: \[\left( \frac{- 8 + 16}{2}, \frac{- 1 - 1}{2} \right) = \left( 4, - 1 \right)\] If the other focus is  \[S'\left( m, n \right)\] ,then it is calculated in the following way:

\[4 = \frac{17 + m}{2}\]

\[ \Rightarrow m = - 9\]

And \[- 1 = \frac{- 1 + n}{2}\]

\[ \Rightarrow n = - 1\]

Thus, the other focus is  \[\left( - 9, - 1 \right)\].

Distance between the vertices: 

\[ 2a = \sqrt{\left( 16 + 8 \right)^2 + \left( - 1 + 1 \right)^2}\]

\[ \Rightarrow 2a = 24\]

\[ \Rightarrow a = 12\]

Distance between the foci:

\[2c = \sqrt{\left( 17 + 9 \right)^2 + \left( - 1 + 1 \right)^2}\]

\[ \Rightarrow 2c = 26\]

\[ \Rightarrow c = 13\]

\[Also, c^2 = a^2 + b^2 \]

\[ \Rightarrow b^2 = 169 - 144\]

\[ \Rightarrow b^2 = 25\]

Equation of the hyperbola is given below:

\[\frac{\left( x - 4 \right)^2}{144} - \frac{\left( y + 1 \right)^2}{25} = 1\]

\[ \Rightarrow \frac{x^2 - 8x + 16}{144} - \frac{\left( y^2 + 2y + 1 \right)}{25} = 1\]

\[ \Rightarrow 25 x^2 - 200x + 400 - \left( 144 y^2 + 288y + 144 \right) = 3600\]

\[ \Rightarrow 25 x^2 - 200x - 144 y^2 - 288y - 3344 = 0\]