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JEE Mathematics 2024 Question with Solution

Let the foci and length of the latus rectum of an ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a>ba > b be (±5,0)(\pm 5, 0) and 50\sqrt{50}, respectively. Then, the square of the eccentricity of the hyperbola x2b2−y2a2b2=1\frac{x^2}{b^2} - \frac{y^2}{a^2b^2} = 1 equals:

  • A

    5151

  • B

    4848

  • C

    5050

  • D

    4545

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The ellipse is x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 with foci (±5,0)(\pm 5,0) and latus rectum length 50\sqrt{50}.

Find: The square of the eccentricity of the hyperbola x2b2−y2a2b2=1\frac{x^2}{b^2} - \frac{y^2}{a^2b^2} = 1.

For the ellipse, the focal distance is ae=5ae = 5 and the length of latus rectum is

2b2a=50\frac{2b^2}{a} = \sqrt{50}

Also,

b2=a2(1−e2)b^2 = a^2(1-e^2)

Using 2b2a=50\frac{2b^2}{a} = \sqrt{50},

b2=a502=52a2b^2 = \frac{a\sqrt{50}}{2} = \frac{5\sqrt{2}a}{2}

Now substitute in b2=a2(1−e2)b^2 = a^2(1-e^2):

a2(1−e2)=52a2a^2(1-e^2) = \frac{5\sqrt{2}a}{2}

So,

a(1−e2)=522a(1-e^2) = \frac{5\sqrt{2}}{2}

Since ae=5ae = 5, we have a=5ea = \frac{5}{e}. Substituting,

5e(1−e2)=522\frac{5}{e}(1-e^2) = \frac{5\sqrt{2}}{2}

Hence,

2−2e2=e\sqrt{2} - \sqrt{2}e^2 = e

which gives

2e2+e−2=0\sqrt{2}e^2 + e - \sqrt{2} = 0

Factorizing,

(e+2)(2e−1)=0(e+\sqrt{2})(\sqrt{2}e-1)=0

So the admissible value is

e=12e = \frac{1}{\sqrt{2}}

Therefore,

a=5e=52a = \frac{5}{e} = 5\sqrt{2}

and hence

a2=50a^2 = 50

For the hyperbola x2b2−y2a2b2=1\frac{x^2}{b^2} - \frac{y^2}{a^2b^2} = 1, comparing with the standard form x2A2−y2B2=1\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1, we get

A2=b2,B2=a2b2A^2 = b^2, \qquad B^2 = a^2b^2

Therefore, its eccentricity satisfies

eH2=1+B2A2=1+a2b2b2=1+a2=51e_H^2 = 1 + \frac{B^2}{A^2} = 1 + \frac{a^2b^2}{b^2} = 1+a^2 = 51

Thus, the correct option is A.

Using ellipse relations first

From the solution working,

foci≡(±5,0),2b2a=50\text{foci} \equiv (\pm 5,0), \qquad \frac{2b^2}{a} = \sqrt{50}

For the ellipse,

ae=5,b2=a2(1−e2)ae = 5, \qquad b^2 = a^2(1-e^2)

and also

b2=52a2b^2 = \frac{5\sqrt{2}a}{2}

Hence,

a(1−e2)=522a(1-e^2) = \frac{5\sqrt{2}}{2}

Using a=5ea = \frac{5}{e},

5e(1−e2)=522\frac{5}{e}(1-e^2) = \frac{5\sqrt{2}}{2}

which leads to

2e2+e−2=0\sqrt{2}e^2 + e - \sqrt{2} = 0

So,

(e+2)(2e−1)=0(e+\sqrt{2})(\sqrt{2}e-1)=0

Thus,

e=−2  (rejected),e=12e = -\sqrt{2} \; (\text{rejected}), \qquad e = \frac{1}{\sqrt{2}}

Then

a=52a = 5\sqrt{2}

For the hyperbola,

eH2=1+a2=1+50=51e_H^2 = 1 + a^2 = 1 + 50 = 51

Therefore, the square of the eccentricity is 5151.

Common mistakes

  • Using the hyperbola eccentricity formula directly with the ellipse parameters without first identifying that, for x2b2−y2a2b2=1\frac{x^2}{b^2} - \frac{y^2}{a^2b^2} = 1, the hyperbola has A2=b2A^2=b^2 and B2=a2b2B^2=a^2b^2. This is wrong because the symbols aa and bb in the hyperbola are not in standard-position roles. First compare with standard form, then use eH2=1+B2A2e_H^2 = 1 + \frac{B^2}{A^2}.

  • Taking the root e=−2e=-\sqrt{2} or e=2e=\sqrt{2} for the ellipse. This is wrong because an ellipse must satisfy 00

  • Misusing the latus rectum formula as 2a2b\frac{2a^2}{b} instead of 2b2a\frac{2b^2}{a} for the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 with major axis along the xx-axis. Use the correct relation before substituting into b2=a2(1−e2)b^2=a^2(1-e^2).

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