A hyperbola, having the transverse axis of length 2 sin , is confocal with the ellipse 3 x 2 + 4 y 2 = 12 . Then its equation is [2007]

Question

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 [2007]

Smart Achievers · Accepted Answer

(1)

$The length of transverse axis = 2 \sin θ = 2 a$

$\Rightarrow a = \sin θ$

$Also the given ellipse is 3 x^{2} + 4 y^{2} = 12$

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

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

$∴$ $Focus of ellipse = (2 \times \frac{1}{2}, 0) \Rightarrow (1, 0)$

$Since hyperbola is confocal with ellipse, therefore focus of hyperbola$

$= (1, 0) \Rightarrow a e = 1 \Rightarrow \sin θ \times e = 1$

$\Rightarrow e = c o s e c θ$

$∴$ $b^{2} = a^{2} (e^{2} - 1)$ $= \sin^{2} θ (\cos e c^{2} θ - 1) = \cos^{2} θ$

$∴$ $Equation of hyperbola is$

$\frac{x^{2}}{\sin^{2} θ} - \frac{y^{2}}{\cos^{2} θ} = 1$

$\Rightarrow x^{2} c o s e c^{2} θ - y^{2} \sec^{2} θ = 1$

Solution

Q.
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 [2007]

1 $x^{2} c o s e c^{2} θ - y^{2} \sec^{2} θ = 1$

2 $x^{2} \sec^{2} θ - y^{2} c o s e c^{2} θ = 1$

3 $x^{2} \sin^{2} θ - y^{2} \cos^{2} θ = 1$

4 $x^{2} \cos^{2} θ - y^{2} \sin^{2} θ = 1$

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Solution

Q. A hyperbola, having the transverse axis of length 2sinθ, is confocal with the ellipse 3x2+4y2=12. Then its equation is [2007]

1 x2cosec2θ-y2sec2θ=1

2 x2sec2θ-y2cosec2θ=1

3 x2sin2θ-y2cos2θ=1