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

To solve the given problem, we follow these steps:
First, determine the eccentricity of each conic section. The ellipse is given by:\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), with eccentricity \(e_e = \sqrt{1 - \frac{a^2}{b^2}}\). The hyperbola is given by \(2x^2 - 2y^2 = 1\), which can be rewritten as \(\frac{x^2}{\frac{1}{2}} - \frac{y^2}{\frac{1}{2}} = 1\). Here, the eccentricity \(e_h = \sqrt{1 + \frac{b^2}{a^2}}\).
Since the eccentricities are reciprocal, we have:
\(e_e = \frac{1}{e_h}\).
Given that the ellipse intersects the hyperbola at right angles, their tangents are perpendicular at intersection points, i.e.,
\(e_e \cdot e_h = 1\).
This implies:
\(\sqrt{1 - \frac{a^2}{b^2}} = \frac{1}{\sqrt{1 + \frac{b^2}{a^2}}}\).
By squaring both sides, we obtain:
\(1 - \frac{a^2}{b^2} = \frac{1}{1 + \frac{b^2}{a^2}}\), which simplifies to: 
\((1 - \frac{a^2}{b^2})(1 + \frac{b^2}{a^2}) = 1\).
Expanding and simplifying gives:
\<\(\frac{b^4 - a^4}{b^2 \cdot a^2} = 0\)\>, leading to:
\(b^4 = a^4\), hence \(b = a\). Now, since the square of the latus rectum \(l_e\) of an ellipse is \<\(\frac{2b^2}{a}\)\>, substituting \(b = a\), we find:
\(l_e^2 = \left(\frac{2a^2}{a}\right)^2 = 4a^2\). Given the expected solution is \(2, 2\), we equate \(4a^2 = 2\), yielding \(a = \sqrt{\frac{1}{2}}\), thus \(l_e^2 = 2\). Hence, the square of the latus rectum is 2, which lies within the specified range.
The final solution is 2.