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

$(2007, 3 M)$

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

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

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Answer:

Correct Answer: 11. (b)

Solution:

The given ellipse is

$\begin{aligned} \frac{x^{2}}{4} + \frac{y^{2}}{3} & = 1 \Rightarrow a = 2, b = \sqrt{3} \\ ∴ & 3 & = 4 (1 - e^{2}) \Rightarrow e = \frac{1}{2} \\ So, & a & = 2 \times \frac{1}{2} = 1 \end{aligned}$

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

$\begin{aligned} e_{1} & = cosec θ [∵ a e = e \sin θ] \\ \Rightarrow & b^{2} & = \sin^{2} θ ({cosec}^{2} θ - 1) & = \cos^{2} θ \end{aligned}$

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. 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

Answer:

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Hyperbola 1 Question 10

11. A hyperbola, having the transverse axis of length 2sin⁡θ, is confocal with the ellipse 3x2+4y2=12. Then, its equation is

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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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