Question

Find the equation of the hyperbola whose vertices are at (± 6, 0) and one of the directrices is x = 4.

Answer 1

The Vertices of the hyperbola are \[\left( \pm 6, 0 \right)\]

∴ \[a = 6\] 

⇒ a² = 36
Now, x = 4

\[\frac{a}{e} = 4\]

\[ \Rightarrow e = \frac{3}{2} \left[ \because a = 6 \right]\]

Now,

\[\left( ae \right)^2 = a^2 + b^2 \]

\[ \Rightarrow \left( 6 \times \frac{3}{2} \right)^2 = 6^2 + b^2 \]

\[ \Rightarrow 81 - 36 = b^2 \]

\[ \Rightarrow b^2 = 45\]

Therefore, the equation of the hyperbola is \[\frac{x^2}{36} - \frac{y^2}{45} = 1\].