JEE Challenger
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MathMedAdv 2009 • Paper 2

Orthogonal Intersection of Ellipse and Hyperbola with Reciprocal Eccentricity

An ellipse intersects the hyperbola 2x2−2y2=12x^2 - 2y^2 = 1 orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then

Options

A

Equation of ellipse is x2+2y2=2x^2 + 2y^2 = 2

Correct
B

The foci of ellipse are (±1,0)(\pm 1, 0)

Correct
C

Equation of ellipse is x2+2y2=4x^2 + 2y^2 = 4

D

The foci of ellipse are (±2,0)(\pm\sqrt{2}, 0)

Topics & Concepts

Conic SectionsEllipseHyperbola

Step-by-Step Solution

To determine the equation and foci of the ellipse, we analyze the given conditions step-by-step:

1. Characteristics of the Hyperbola:
The given equation of the hyperbola is:
2x2−2y2=1  ⟹  x21/2−y21/2=12x^2 - 2y^2 = 1 \implies \frac{x^2}{1/2} - \frac{y^2}{1/2} = 1
This is a rectangular hyperbola where:
aH2=12,bH2=12a_H^2 = \frac{1}{2}, \quad b_H^2 = \frac{1}{2}
The eccentricity eHe_H of this hyperbola is:
eH=1+bH2aH2=1+1=2e_H = \sqrt{1 + \frac{b_H^2}{a_H^2}} = \sqrt{1 + 1} = \sqrt{2}
The distance of each focus from the origin, cc, satisfies:
c2=aH2+bH2=12+12=1  ⟹  c=1c^2 = a_H^2 + b_H^2 = \frac{1}{2} + \frac{1}{2} = 1 \implies c = 1
Thus, the foci of the hyperbola are at (±1,0)(\pm 1, 0).

2. Eccentricity of the Ellipse:
The eccentricity eEe_E of the ellipse is the reciprocal of the hyperbola's eccentricity:
eE=1eH=12e_E = \frac{1}{e_H} = \frac{1}{\sqrt{2}}

3. Orthogonal Intersection and Confocal Property:
Let the equation of the ellipse be:
x2a2+y2b2=1(a>b)\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (a > b)
Differentiating the ellipse with respect to xx:
2xa2+2yb2dydx=0  ⟹  m1=−b2xa2y\frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0 \implies m_1 = -\frac{b^2 x}{a^2 y}
Differentiating the hyperbola x2aH2−y2bH2=1\frac{x^2}{a_H^2} - \frac{y^2}{b_H^2} = 1 with respect to xx:
2xaH2−2ybH2dydx=0  ⟹  m2=bH2xaH2y\frac{2x}{a_H^2} - \frac{2y}{b_H^2}\frac{dy}{dx} = 0 \implies m_2 = \frac{b_H^2 x}{a_H^2 y}
For orthogonal intersection, m1m2=−1m_1 m_2 = -1:
(−b2xa2y)(bH2xaH2y)=−1  ⟹  x2y2=a2aH2b2bH2\left(-\frac{b^2 x}{a^2 y}\right)\left(\frac{b_H^2 x}{a_H^2 y}\right) = -1 \implies \frac{x^2}{y^2} = \frac{a^2 a_H^2}{b^2 b_H^2}
Subtracting the equations of the two conics at their point of intersection gives:
x2(1a2−1aH2)+y2(1b2+1bH2)=0  ⟹  x2y2=1b2+1bH21aH2−1a2=a2aH2(b2+bH2)b2bH2(a2−aH2)x^2\left(\frac{1}{a^2} - \frac{1}{a_H^2}\right) + y^2\left(\frac{1}{b^2} + \frac{1}{b_H^2}\right) = 0 \implies \frac{x^2}{y^2} = \frac{\frac{1}{b^2} + \frac{1}{b_H^2}}{\frac{1}{a_H^2} - \frac{1}{a^2}} = \frac{a^2 a_H^2 (b^2 + b_H^2)}{b^2 b_H^2 (a^2 - a_H^2)}
Equating the two expressions for x2y2\frac{x^2}{y^2}:
b2+bH2a2−aH2=1  ⟹  a2−b2=aH2+bH2\frac{b^2 + b_H^2}{a^2 - a_H^2} = 1 \implies a^2 - b^2 = a_H^2 + b_H^2
This proves that the two conics are confocal, meaning they share the same foci:
c2=a2−b2=aH2+bH2=1c^2 = a^2 - b^2 = a_H^2 + b_H^2 = 1

Therefore:

  • The foci of the ellipse are (±c,0)=(±1,0)(\pm c, 0) = (\pm 1, 0).

4. Equation of the Ellipse:
Using the eccentricity of the ellipse:
eE=ca  ⟹  12=1a  ⟹  a=2  ⟹  a2=2e_E = \frac{c}{a} \implies \frac{1}{\sqrt{2}} = \frac{1}{a} \implies a = \sqrt{2} \implies a^2 = 2
Since a2−b2=1a^2 - b^2 = 1:
b2=a2−1=2−1=1b^2 = a^2 - 1 = 2 - 1 = 1
Hence, the equation of the ellipse is:
x22+y21=1  ⟹  x2+2y2=2\frac{x^2}{2} + \frac{y^2}{1} = 1 \implies x^2 + 2y^2 = 2

Thus, statements (A) and (B) are correct.

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