SATHEE: Hyperbola 1 Question 7

Hyperbola 1 Question 7

8.

Let $0 < θ < \frac{π}{2}$ . If the eccentricity of the hyperbola $\frac{x^{2}}{\cos^{2} θ} - \frac{y^{2}}{\sin^{2} θ} = 1$ is greater than 2 , then the length of its latus rectum lies in the interval

(2019 Main, 9 Jan I)

(a) $1, \frac{3}{2}$

(b) $(3, \infty)$

(d) $(2, 3)$

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Answer:

Correct Answer: 8. (b)

Solution:

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

$e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$

$∴$ For the given hyperbola,

$e = \sqrt{1 + \frac{\sin^{2} θ}{\cos^{2} θ}} > 2$

$\Rightarrow 1 + \tan^{2} θ > 4$

$(∵ a^{2} = \cos^{2} θ and b^{2} = \sin^{2} θ)$

$\Rightarrow \tan^{2} θ > 3$

$\Rightarrow \tan θ \in (- \infty, - \sqrt{3}) \cup (\sqrt{3}, \infty)$

$[x^{2} > 3 \Rightarrow | x | > \sqrt{3} \Rightarrow x \in (- \infty, - \sqrt{3}) \cup (\sqrt{3}, \infty)]$

But $θ \in 0, \frac{π}{2} \Rightarrow \tan θ \in (\sqrt{3}, \infty)$

$\Rightarrow θ \in \frac{π}{3}, \frac{π}{2}$

Now, length of latusrectum

$= \frac{2 b^{2}}{a} = 2 \frac{\sin^{2} θ}{\cos θ} = 2 \sin θ \tan θ$

Since, both $\sin θ$ and $\tan θ$ are increasing functions in $\frac{π}{3}, \frac{π}{2}$ .

$∴$ Least value of latusrectum is

$= 2 \sin \frac{π}{3} \cdot \tan \frac{π}{3} = 2 \cdot \frac{\sqrt{3}}{2} \cdot \sqrt{3} = 3 at θ = \frac{π}{3}$

and greatest value of latusrectum is $< \infty$

Hence, latusrectum length $\in (3, \infty)$

Hyperbola 1 Question 7

8.

Answer:

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Hyperbola 1 Question 7

8.

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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