SATHEE: Hyperbola 1 Question 14

Hyperbola 1 Question 14

15.

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be reciprocal to that of the ellipse $x^{2} + 4 y^{2} = 4$ . If the hyperbola passes through a focus of the ellipse, then

(a) the equation of the hyperbola is $\frac{x^{2}}{3} - \frac{y^{2}}{2} = 1$

(b) a focus of the hyperbola is $(2, 0)$

(d) the equation of the hyperbola is $x^{2} - 3 y^{2} = 3$

(2011)

Show Answer

Answer:

Correct Answer: 15. (b, d)

Solution:

Here, equation of ellipse is $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$

$\Rightarrow e^{2} = 1 - \frac{b^{2}}{a^{2}} = 1 - \frac{1}{4} = \frac{3}{4}$

$∴ e = \frac{\sqrt{3}}{2}$ and focus $(\pm a e, 0) \Rightarrow (\pm \sqrt{3}, 0)$

For hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, e_{1}^{2} = 1 + \frac{b^{2}}{a^{2}}$

where,

$e_{1}^{2} = \frac{1}{e^{2}} = \frac{4}{3} \Rightarrow 1 + \frac{b^{2}}{a^{2}} = \frac{4}{3}$

$∴ \frac{b^{2}}{a^{2}} = \frac{1}{3}$

and hyperbola passes through $(\pm \sqrt{3}, 0)$

$\Rightarrow \frac{3}{a^{2}} = 1 \Rightarrow a^{2} = 3$

From Eqs. (i) and (ii), $b^{2} = 1$

$∴$ Equation of hyperbola is $\frac{x^{2}}{3} - \frac{y^{2}}{1} = 1$

Focus is $(\pm a e_{1}, 0) \Rightarrow \pm \sqrt{3} \cdot \frac{2}{\sqrt{3}}, 0 \Rightarrow (\pm 2, 0)$

Hence, (b) and (d) are correct answers.

Hyperbola 1 Question 14

15.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Hyperbola 1 Question 14

15.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu