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) $(5,0)$

Show Answer

Answer:

Correct Answer: 2. (b)

Solution:

  1. 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 $-3 \frac{5}{3}, 0=(-5,0)$



NCERT Chapter Video Solution

Dual Pane