Hyperbola 1 Question 1
2. If $5 x+9=0$ is the directrix of the hyperbola $16 x^{2}-9 y^{2}=144$, then its corresponding focus is
(2019 Main, 10 April II)
(a) $-\frac{5}{3}, 0$
(b) $(-5,0)$
(c) $\frac{5}{3}, 0$
(d) $(5,0)$
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Answer:
Correct Answer: 2. (b)
Solution:
- Equation of given hyperbola is
$$ \begin{aligned} & 16 x^{2}-9 y^{2}=144 \\ & \Rightarrow \quad \frac{x^{2}}{9}-\frac{y^{2}}{16}=1 \end{aligned} $$
So, the eccentricity of Eq. (i)
$$ e=\sqrt{1+\frac{16}{9}}=\frac{5}{3} $$
[ $\because$ the eccentricity (e) of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $\left.\sqrt{1+(b / a)^{2}}\right]$ and given directrix is $5 x+9=0 \Rightarrow x=-9 / 5$
So, corresponding focus is $-3 \frac{5}{3}, 0=(-5,0)$
