SATHEE: Hyperbola 1 Question 10

Hyperbola 1 Question 10

11.

A hyperbola, having the transverse axis of length $2 \sin θ$ , is confocal with the ellipse $3 x^{2} + 4 y^{2} = 12$ . Then, its equation is

(a) $x^{2} {cosec}^{2} θ - y^{2} \sec^{2} θ = 1$

(b) $x^{2} \sec^{2} θ - y^{2} {cosec}^{2} θ = 1$

(d) $x^{2} \cos^{2} θ - y^{2} \sin^{2} θ = 1$

$(2007, 3 M)$

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Answer:

Correct Answer: 11. (a)

Solution:

The given ellipse is

$\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1 \Rightarrow a = 2, b = \sqrt{3}$

$∴ 3 = 4 (1 - e^{2}) \Rightarrow e = \frac{1}{2}$

$So, a = 2 \times \frac{1}{2} = 1$

Hence, the eccentricity $e_{1}$ of the hyperbola is given by

$e_{1} = cosec θ [∵ a e = e \sin θ]$

$\Rightarrow b^{2} = \sin^{2} θ ({cosec}^{2} θ - 1) = \cos^{2} θ$

Hence, equation of hyperbola is

$\frac{x^{2}}{\sin^{2} θ} - \frac{y^{2}}{\cos^{2} θ} = 1$ or $x^{2} {cosec}^{2} θ - y^{2} \sec^{2} θ = 1$ .

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Hyperbola 1 Question 10

11.

Answer:

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