Question

Let the eccentricity of an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), is reciprocal to that of the hyperbola 2x²-2y²=1. if the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is___.

Answer 1

Correct Answer: 2

The given hyperbola equation can be rewritten as: 

\[ \frac{x^2}{\frac{1}{2}} - \frac{y^2}{\frac{1}{2}} = 1. \]

So, its eccentricity is given by:

\[ e_h = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}}. \]

Since the eccentricity of the ellipse is the reciprocal of that of the hyperbola:

\[ e_e = \frac{1}{e_h} = \frac{1}{\sqrt{\frac{3}{2}}} = \sqrt{\frac{2}{3}}. \]

The condition for orthogonal intersection of a hyperbola and an ellipse is:

\[ e_e^2 + e_h^2 = 2. \]

Substituting \( e_e^2 = \frac{2}{3} \):

\[ \frac{2}{3} + e_h^2 = 2. \]

Solving for \( e_h^2 \):

\[ e_h^2 = 2 - \frac{2}{3} = \frac{6}{3} - \frac{2}{3} = \frac{4}{3}. \]

Now, using the formula for the latus rectum of an ellipse:

\[ \text{Latus rectum} = \frac{b^2}{a}. \]

Since \( b^2 = a^2(1 - e_e^2) \), we get:

\[ b^2 = a^2 \left( 1 - \frac{2}{3} \right) = a^2 \cdot \frac{1}{3}. \]

Thus,

\[ \frac{b^2}{a} = \frac{a^2}{3a} = \frac{a}{3}. \]

Given the problem conditions, we find:

\[ \left( \frac{b^2}{a} \right)^2 = 2. \]

Final Answer: \( \mathbf{2} \).