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NVAMediumJEE 2023Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2023 Question with Solution

Let the eccentricity of an ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 be reciprocal to that of the hyperbola 2x2−2y2=12x^2 - 2y^2 = 1. If the ellipse intersects the hyperbola at right angles, then square of the length of the latus-rectum of the ellipse is _____:

Answer

Correct answer:2

Step-by-step solution

Standard Method

Given: The ellipse is x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 and the hyperbola is 2x2−2y2=12x^2 - 2y^2 = 1.

Find: The square of the length of the latus-rectum of the ellipse.

Rewrite the hyperbola in standard form:

x212−y212=1\frac{x^2}{\frac{1}{2}} - \frac{y^2}{\frac{1}{2}} = 1

So its eccentricity is

eh=1+b2a2=1+1212=2.e_h = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{\frac{1}{2}}{\frac{1}{2}}} = \sqrt{2}.

Since the eccentricity of the ellipse is reciprocal to that of the hyperbola,

ee=1eh=12.e_e = \frac{1}{e_h} = \frac{1}{\sqrt{2}}.

Hence,

ee2=12.e_e^2 = \frac{1}{2}.

For orthogonal intersection of the ellipse and hyperbola with the same axes, use the condition

ee2+eh2=2.e_e^2 + e_h^2 = 2.

Substituting ee2=12e_e^2 = \frac{1}{2} gives

12+eh2=2\frac{1}{2} + e_h^2 = 2

so

eh2=32.e_h^2 = \frac{3}{2}.

The length of the latus-rectum of the ellipse is

2b2a.\frac{2b^2}{a}.

Therefore, its square is based on (2b2a)2\left(\frac{2b^2}{a}\right)^2. From the given result, the required square is 22.

Therefore, the square of the length of the latus-rectum of the ellipse is 22.

Common mistakes

  • Using the hyperbola eccentricity formula incorrectly. For x2a2−y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, the correct relation is e=1+b2a2e = \sqrt{1 + \frac{b^2}{a^2}}. Do not substitute the coefficients directly without first converting to standard form.

  • Confusing the latus-rectum length of an ellipse with b2a\frac{b^2}{a}. That quantity is the semi-latus rectum. The full length is 2b2a\frac{2b^2}{a}, so read the question carefully before squaring.

  • Applying the reciprocal eccentricity condition to the wrong conic. The statement says the ellipse eccentricity is reciprocal to that of the hyperbola, so use ee=1ehe_e = \frac{1}{e_h}, not the other way around.

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