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:

• A. \(x^2+2y^2=4\)
• B. \(2x^2+y^2=4\)
• C. \(2x^2+y^2=2\)
• D. \(x^2+2y^2=2\)

Hint:

Always convert conics into standard form before using eccentricity formulas.

Answer 1

The Correct Option is D

Step 1: Write the hyperbola neatly.
$2x^2-2y^2=1$ becomes $\dfrac{x^2}{1/2}-\dfrac{y^2}{1/2}=1$, so $a^2=b^2=\dfrac12$.
Step 2: Find the eccentricity of the hyperbola.
\[ e_h=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+1}=\sqrt2. \]
Step 3: Get the ellipse eccentricity.
It is the reciprocal, so $e_e=\dfrac{1}{\sqrt2}$.
Step 4: Use the ellipse eccentricity formula.
For $\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1$, $e_e=\sqrt{1-\dfrac{B^2}{A^2}}$. So \[ 1-\frac{B^2}{A^2}=\frac12\Rightarrow \frac{B^2}{A^2}=\frac12\Rightarrow A^2=2B^2. \]
Step 5: Test the options against $A^2=2B^2$.
Option (D) $x^2+2y^2=2$ is $\dfrac{x^2}{2}+\dfrac{y^2}{1}=1$, so $A^2=2$, $B^2=1$, and indeed $A^2=2B^2$.
Step 6: Confirm.
The ratio $\dfrac{B^2}{A^2}=\dfrac12$ matches, so this ellipse fits all the conditions. \[ \boxed{x^2+2y^2=2} \]