SATHEE: Hyperbola 1 Question 15

Hyperbola 1 Question 15

16.

An ellipse intersects the hyperbola $2 x^{2} - 2 y^{2} = 1$ orthogonally. The eccentricity of the ellipse is reciprocal to that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then

(2009)

(a) equation of ellipse is $x^{2} + 2 y^{2} = 2$

(b) the foci of ellipse are $(\pm 1, 0)$

(d) the foci of ellipse are $(\pm \sqrt{2}, 0)$

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Answer:

Correct Answer: 16. (a,b)

Solution:

Given,

$2 x^{2} - 2 y^{2} = 1$

$\Rightarrow \frac{x^{2}}{\frac{1}{2}} - \frac{y^{2}}{\frac{1}{2}} = 1$

Eccentricity of hyperbola $= \sqrt{2}$

So, eccentricity of ellipse $= 1 / \sqrt{2}$

Let equation of ellipse be

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

[where $a > b$ ]

$\begin{aligned} ∴ & \frac{1}{\sqrt{2}} & = \sqrt{1 - \frac{b^{2}}{a^{2}}} \\ \Rightarrow & \frac{b^{2}}{a^{2}} & = \frac{1}{2} \Rightarrow a^{2} = 2 b^{2} \\ \Rightarrow & x^{2} + 2 y^{2} & = 2 b^{2} \end{aligned}$

Let ellipse and hyperbola intersect at

$A \frac{1}{\sqrt{2}} \sec θ, \frac{1}{\sqrt{2}} \tan θ$

On differentiating Eq. (i), we get

$\begin{aligned} 4 x - 4 y \frac{d y}{d x} & = 0 \Rightarrow \frac{d y}{d x} = \frac{x}{y} \\ {\frac{d y}{d x} |}_{at A} & = \frac{\sec θ}{\tan θ} = cosec θ \end{aligned}$

and on differentiating Eq. (ii), we get

$2 x + 4 y \frac{d y}{d x} = {0 \Rightarrow \frac{d y}{d x} |}_{at A} = - \frac{x}{2 y} = - \frac{1}{2} cosec θ$

Since, ellipse and hyperbola are orthogonal.

$∴ - \frac{1}{2} {cosec}^{2} θ = - 1 \Rightarrow {cosec}^{2} θ = 2 \Rightarrow θ = \pm \frac{π}{4}$

$∴ A 1, \frac{1}{\sqrt{2}}$ or $1, - \frac{1}{\sqrt{2}}$

Form Eq. (ii), $1 + 2 {\frac{1}{\sqrt{2}}}^{2} = 2 b^{2}$

$\Rightarrow b^{2} = 1$

Equation of ellipse is $x^{2} + 2 y^{2} = 2$ .

Coordinates of foci $(\pm a e, 0) = \pm \sqrt{2} \cdot \frac{1}{\sqrt{2}}, 0 = (\pm 1, 0)$

If major axis is along $Y$ -axis, then

$\begin{aligned} \frac{1}{\sqrt{2}} & = \sqrt{1 - \frac{a^{2}}{b^{2}}} \Rightarrow b^{2} = 2 a^{2} \\ ∴ 2 x^{2} + y^{2} & = 2 a^{2} \Rightarrow Y^{'} = - \frac{2 x}{y} \\ \Rightarrow y_{\frac{1}{\sqrt{2}} \sec θ, \frac{1}{\sqrt{2}} \tan θ} & = \frac{- 2}{\sin θ} \end{aligned}$

As ellipse and hyperbola are orthogonal

$\begin{aligned} ∴ - \frac{2}{\sin θ} \cdot cosec θ = - 1 \\ \Rightarrow {cosec}^{2} θ = \frac{1}{2} \Rightarrow θ = \pm \frac{π}{4} \\ ∴ 2 x^{2} + y^{2} = 2 a^{2} \\ \Rightarrow 2 + \frac{1}{2} = 2 a^{2} \Rightarrow a^{2} = \frac{5}{4} \\ ∴ 2 x^{2} + y^{2} = \frac{5}{2}, corresponding foci are (0, \pm 1) \end{aligned}$

Hence, option (a) and (b) are correct.

Hyperbola 1 Question 15

16.

Answer:

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Hyperbola 1 Question 15

16.

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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