SATHEE: Hyperbola 1 Question 1

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) $(- 5, 0)$

(d) $(5, 0)$

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Answer:

Correct Answer: 2. (b)

Solution:

Equation of given hyperbola is

$\begin{aligned} 16 x^{2} - 9 y^{2} = 144 \\ \Rightarrow \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1 \end{aligned}$

So, the eccentricity of Eq. (i)

$e = \sqrt{1 + \frac{16}{9}} = \frac{5}{3}$

[ $∵$ the eccentricity (e) of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $\sqrt{1 + (b / a)^{2}}]$ and given directrix is $5 x + 9 = 0 \Rightarrow x = - 9 / 5$

So, corresponding focus is $- 3 \frac{5}{3}, 0 = (- 5, 0)$

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

Answer:

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Hyperbola 1 Question 1

2. If 5x+9=0 is the directrix of the hyperbola 16x2−9y2=144, then its corresponding focus is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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2. If $5 x + 9 = 0$ is the directrix of the hyperbola $16 x^{2} - 9 y^{2} = 144$ , then its corresponding focus is