JEE PYQ: Hyperbola Question 13
Question 13 - 2019 (10 Apr Shift 1)
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:
(1) $4e^4 - 24e^2 + 27 = 0$
(2) $4e^4 - 12e^2 - 27 = 0$
(3) $4e^4 - 24e^2 + 35 = 0$
(4) $4e^4 + 8e^2 - 35 = 0$
Show Answer
Answer: (3)
Solution
Directrix: $x = \frac{4\sqrt{5}}{5} = \frac{4}{\sqrt{5}}$, so $\frac{a}{e} = \frac{4}{\sqrt{5}}$. Hyperbola passes through $(4, -2\sqrt{3})$: $\frac{16}{a^2} - \frac{12}{a^2(e^2-1)} = 1$. From $a = \frac{4e}{\sqrt{5}}$: $a^2 = \frac{16e^2}{5}$. Substituting: $\frac{16 \cdot 5}{16e^2} - \frac{12 \cdot 5}{16e^2(e^2-1)} = 1$, so $\frac{5}{e^2} - \frac{15}{4e^2(e^2-1)} = 1$. Simplifying: $\frac{4(e^2-1)\cdot 5 - 15}{4e^2(e^2-1)} = 1$, giving $20e^2 - 35 = 4e^4 - 4e^2$. Hence $4e^4 - 24e^2 + 35 = 0$.
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