Let the eccentricity of an ellipse x²/a²+y²/b²=1 is reciprocal to that of the hyperbola 2 x²-2 y²=1. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is ___________.

Solution

Maths sir · 4 steps

Get b² in terms of a² from the eccentricity, then let the right-angle condition fix a².

1. Fix the eccentricity. The hyperbola is x² - y² = 1/2. It is rectangular, so its eccentricity is √2. The ellipse then has e = 1/√2, and b² = a²(1 - e²) gives b² = a²/2.
2. Slopes at meeting point. At a common point (x, y), the ellipse has slope -(b² x)/(a² y) and the hyperbola has slope x/y. The curves meet at right angles, so the product of the slopes is -1.

   (-(b² x)/(a² y))(x/y) = -1 ⇒ b²x² = a²y² ⇒ y² = x²/2

3. Find the point. Put y² = x²/2 into 2x² - 2y² = 1. This gives x² = 1 and y² = 1/2. The point is also on the ellipse x² + 2y² = a², so a² = 1 + 1 = 2 and b² = 1.
4. So. The latus rectum is 2b²/a = 2/√2 = √2. Its square is 2.

The common slip: 1.41. Reporting the latus rectum itself, √2, gives about 1.41. The question asks for its square, which is 2.

The idea to keep. A standard ellipse and hyperbola cut at right angles exactly when they share foci: a² - b² = A² + B². Here A² + B² = 1 gives a² - b² = 1 straight away.

Answer from the NTA final answer key. Solution by PrepWiser.