Question

If a directrix of a hyperbola centred at the origin and passing through the point \((4, -2\sqrt{3})\) is \(5x = 4\sqrt{5}\) and its eccentricity is \(e\), then :

• A. \(4e^4 - 24e^2 + 27 = 0\)
• B. \(4e^4 - 24e^2 + 35 = 0\)
• C. \(4e^4 - 12e^2 - 27 = 0\)
• D. \(4e^4 + 8e^2 - 35 = 0\)

Hint:

Use directrix \(x = a/e\) to get \(a^2\) in terms of \(e\), use \(b^2 = a^2(e^2-1)\) and substitute the given point into \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\).

Answer 1

The Correct Option is B

Let the hyperbola be \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) with \(b^2 = a^2(e^2-1)\).
Directrix: \(x = \frac{a}{e}\). Given \(x = \frac{4}{\sqrt{5}}\), so \(a = \frac{4e}{\sqrt{5}}\) and \(a^2 = \frac{16e^2}{5}\).
Then \(b^2 = \frac{16e^2(e^2-1)}{5}\).
Point \((4,-2\sqrt{3})\) lies on the curve: \(\frac{16}{a^2}-\frac{12}{b^2}=1\).
\(\frac{16}{a^2} = \frac{5}{e^2}\) and \(\frac{12}{b^2} = \frac{60}{16e^2(e^2-1)} = \frac{15}{4e^2(e^2-1)}\).
So \(\frac{5}{e^2}-\frac{15}{4e^2(e^2-1)} = 1\).
Multiply by \(4e^2(e^2-1)\): \(20(e^2-1) - 15 = 4e^2(e^2-1)\).
\(20e^2 - 35 = 4e^4 - 4e^2\) \(\Rightarrow\) \(4e^4 - 24e^2 + 35 = 0\).
Answer: (2) \(4e^4 - 24e^2 + 35 = 0\)