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प्रश्न
Find the equation of the hyperbola whose vertices are (−8, −1) and (16, −1) and focus is (17, −1).
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उत्तर
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:
\[\frac{\left( x - 4 \right)^2}{144} - \frac{\left( y + 1 \right)^2}{25} = 1\]
\[ \Rightarrow \frac{x^2 - 8x + 16}{144} - \frac{\left( y^2 + 2y + 1 \right)}{25} = 1\]
\[ \Rightarrow 25 x^2 - 200x + 400 - \left( 144 y^2 + 288y + 144 \right) = 3600\]
\[ \Rightarrow 25 x^2 - 200x - 144 y^2 - 288y - 3344 = 0\]
