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MCQMediumJEE 2024Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2024 Question with Solution

If the foci of a hyperbola are the same as that of the ellipse x29+y225=1\frac{x^2}{9} + \frac{y^2}{25} = 1 and the eccentricity of the hyperbola is 158\frac{15}{8} times the eccentricity of the ellipse, then the smaller focal distance is:

  • A

    77

  • B

    1414

  • C

    1414

  • D

    77

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The ellipse is

x29+y225=1\frac{x^2}{9} + \frac{y^2}{25} = 1

and the hyperbola has the same foci as this ellipse. Also, the eccentricity of the hyperbola is 158\frac{15}{8} times the eccentricity of the ellipse.

Find: The smaller focal distance.

For the ellipse,

eellipse=1−925=1625=45e_{\text{ellipse}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}

Hence the foci are

(0,±4)(0, \pm 4)

because the major axis is along the yy-axis.

Therefore the hyperbola also has focal distance

c=4c = 4

Its eccentricity is

ehyperbola=158×45=32e_{\text{hyperbola}} = \frac{15}{8} \times \frac{4}{5} = \frac{3}{2}

For a hyperbola, using

e=cae = \frac{c}{a}

we get

a=ce=43/2=83a = \frac{c}{e} = \frac{4}{3/2} = \frac{8}{3}

The solution concludes that the correct option is A. It also shows a worked value for the smaller focal distance, though the displayed final algebra is inconsistent and appears partially corrupted. Since the solution explicitly states The Correct Option is A, we take the answer as 77.

Therefore, the correct option is A and the smaller focal distance is 77.

Solution

Given: the solution explicitly marks A as the correct option.

Find: The corresponding value.

Option A is

77

while options B and C are 1414 and option D again shows 77.

There is a discrepancy in the listed options because A and D are identical, and the mathematical working in the working is partially malformed. However, the solution directly identifies A as correct.

Therefore, the accepted answer from the provided source is A, i.e. 77.

Common mistakes

  • Taking a=3a = 3 and b=5b = 5 for the ellipse without noticing that the major axis is along the yy-axis. This can lead to wrong focus coordinates. Instead, identify the larger denominator first and place the major axis accordingly.

  • Using the ellipse eccentricity formula for the hyperbola as well. For the hyperbola, use the relation e=cae = \frac{c}{a} with e>1e > 1, not e=1−b2a2e = \sqrt{1 - \frac{b^2}{a^2}}.

  • Ignoring that the hyperbola and ellipse have the same foci. This means the focal distance cc remains the same. Do not recompute a different focus value unless the question states otherwise.

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