JEE PYQ: Hyperbola Question 14
Question 14 - 2019 (10 Apr Shift 2)
If $5x + 9 = 0$ is the directrix of the hyperbola $16x^2 - 9y^2 = 144$, then its corresponding focus is:
(1) $(5, 0)$
(2) $\left(-\frac{5}{3}, 0\right)$
(3) $\left(\frac{5}{3}, 0\right)$
(4) $(-5, 0)$
Show Answer
Answer: (4)
Solution
$\frac{x^2}{9} - \frac{y^2}{16} = 1$, $a = 3$, $b = 4$, $e^2 = 1 + \frac{16}{9} = \frac{25}{9}$, $e = \frac{5}{3}$. Directrix $x = -\frac{a}{e} = -\frac{9}{5}$, which matches $5x + 9 = 0$. The corresponding focus is $S’(-ae, 0) = \left(-3\cdot\frac{5}{3}, 0\right) = (-5, 0)$.
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