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JEE Mathematics 2025 Question with Solution

Let the length of a latus rectum of an ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 be 1010. If its eccentricity is ee, and the minimum value of the function f(t)=t2+t+1112f(t) = t^2 + t + \frac{11}{12}, where t∈Rt \in \mathbb{R}, then a2+b2a^2 + b^2 is equal to:

  • A

    125125

  • B

    126126

  • C

    120120

  • D

    115115

Answer

Correct answer:B

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 length of its latus rectum is 1010.

Find: a2+b2a^2 + b^2.

For the ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, the length of the latus rectum is

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

So,

2b2a=10\frac{2b^2}{a} = 10

which gives

2b2=10a2b^2 = 10a

Hence,

b2=5ab^2 = 5a

Now consider

f(t)=t2+t+1112f(t) = t^2 + t + \frac{11}{12}

Its minimum value occurs at

t=−12t = -\frac{1}{2}

Therefore,

f(−12)=(−12)2+(−12)+1112f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + \frac{11}{12} =14−12+1112= \frac{1}{4} - \frac{1}{2} + \frac{11}{12} =312−612+1112= \frac{3}{12} - \frac{6}{12} + \frac{11}{12} =812=23= \frac{8}{12} = \frac{2}{3}

Using the solution relation that the eccentricity is

e=23e = \frac{2}{3}

For an ellipse,

e2=1−b2a2e^2 = 1 - \frac{b^2}{a^2}

So,

49=1−b2a2\frac{4}{9} = 1 - \frac{b^2}{a^2}

which gives

b2a2=59\frac{b^2}{a^2} = \frac{5}{9}

Now use b2=5ab^2 = 5a:

5aa2=59\frac{5a}{a^2} = \frac{5}{9} 5a=59\frac{5}{a} = \frac{5}{9}

Thus,

a=9a = 9

Then,

b2=5×9=45b^2 = 5 \times 9 = 45

Finally,

a2+b2=92+45=81+45=126a^2 + b^2 = 9^2 + 45 = 81 + 45 = 126

Therefore, the value of a2+b2a^2 + b^2 is 126126, so the correct option is B.

Using derivative to minimize the quadratic

Given: f(t)=t2+t+1112f(t) = t^2 + t + \frac{11}{12}.

Find: The minimum value of the quadratic and then use it in the ellipse relation.

Differentiate:

df(t)dt=2t+1\frac{df(t)}{dt} = 2t + 1

Set the derivative equal to zero:

2t+1=02t + 1 = 0 t=−12t = -\frac{1}{2}

Now substitute this value into the function:

f(−12)=(−12)2+(−12)+1112f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + \frac{11}{12} =14−12+1112=23= \frac{1}{4} - \frac{1}{2} + \frac{11}{12} = \frac{2}{3}

From the provided solution, this minimum value is equated to the eccentricity, so

e=23e = \frac{2}{3}

Using this with the latus rectum condition again leads to a=9a = 9 and b2=45b^2 = 45, hence

a2+b2=126a^2 + b^2 = 126

Thus the correct option is B.

Common mistakes

  • Using the latus rectum formula incorrectly as 2a2b\frac{2a^2}{b}. For the ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, the correct length is 2b2a\frac{2b^2}{a}. Start with the standard formula before substituting values.

  • Finding the minimum of f(t)f(t) incorrectly by substituting an arbitrary value of tt. A quadratic with positive coefficient of t2t^2 attains its minimum at the vertex, here at t=−12t = -\frac{1}{2}, or equivalently by differentiation.

  • Writing the eccentricity relation wrongly as e=1−b2a2e = 1 - \frac{b^2}{a^2}. The correct relation is e2=1−b2a2e^2 = 1 - \frac{b^2}{a^2}. Square the eccentricity before comparing with the ratio of the squares of the axes.

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