Question:
An ellipse intersects the hyperbola \(2x^2-2y^2=1\) 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 the equation of the ellipse is:
An ellipse intersects the hyperbola \(2x^2-2y^2=1\) 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 the equation of the ellipse is:
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Always convert conics into standard form before using eccentricity formulas.
Updated On: Jun 17, 2026
- \(x^2+2y^2=4\)
- \(2x^2+y^2=4\)
- \(2x^2+y^2=2\)
- \(x^2+2y^2=2\)
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The Correct Option is D
Solution and Explanation
Concept:
The standard hyperbola:
\[
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
\]
has eccentricity:
\[
e=\sqrt{1+\frac{b^2}{a^2}}
\]
An ellipse:
\[
\frac{x^2}{A^2}+\frac{y^2}{B^2}=1
\]
has eccentricity:
\[
e=\sqrt{1-\frac{B^2}{A^2}}
\]
Step 1: Find eccentricity of hyperbola.
Given hyperbola: \[ 2x^2-2y^2=1 \] \[ \frac{x^2}{1/2}-\frac{y^2}{1/2}=1 \] Thus, \[ a^2=b^2=\frac12 \] Hence eccentricity: \[ e_h=\sqrt{1+\frac{b^2}{a^2}} =\sqrt{2} \]
Step 2: Eccentricity of ellipse.
Given that ellipse eccentricity is reciprocal: \[ e_e=\frac1{\sqrt2} \] So, \[ 1-\frac{B^2}{A^2}=\frac12 \] \[ \frac{B^2}{A^2}=\frac12 \] \[ A^2=2B^2 \]
Step 3: Check options.
Option (D): \[ x^2+2y^2=2 \] \[ \frac{x^2}{2}+\frac{y^2}{1}=1 \] Thus, \[ A^2=2,\quad B^2=1 \] Indeed, \[ \frac{B^2}{A^2}=\frac12 \] Hence correct option is: \[ \boxed{x^2+2y^2=2} \]
Step 1: Find eccentricity of hyperbola.
Given hyperbola: \[ 2x^2-2y^2=1 \] \[ \frac{x^2}{1/2}-\frac{y^2}{1/2}=1 \] Thus, \[ a^2=b^2=\frac12 \] Hence eccentricity: \[ e_h=\sqrt{1+\frac{b^2}{a^2}} =\sqrt{2} \]
Step 2: Eccentricity of ellipse.
Given that ellipse eccentricity is reciprocal: \[ e_e=\frac1{\sqrt2} \] So, \[ 1-\frac{B^2}{A^2}=\frac12 \] \[ \frac{B^2}{A^2}=\frac12 \] \[ A^2=2B^2 \]
Step 3: Check options.
Option (D): \[ x^2+2y^2=2 \] \[ \frac{x^2}{2}+\frac{y^2}{1}=1 \] Thus, \[ A^2=2,\quad B^2=1 \] Indeed, \[ \frac{B^2}{A^2}=\frac12 \] Hence correct option is: \[ \boxed{x^2+2y^2=2} \]
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