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JEE Mathematics 2026 Question with Solution

Let the length of the latus rectum of an ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 (a>b)(a>b) be 3030. If its eccentricity is the maximum value of the function f(t)=−34+2t−t2f(t)=-\dfrac{3}{4}+2t-t^2, then (a2+b2)(a^2+b^2) is equal to

  • A

    276276

  • B

    516516

  • C

    256256

  • D

    496496

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The ellipse is x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 with a>ba>b, the length of its latus rectum is 3030, and its eccentricity equals the maximum value of f(t)=−t2+2t−34f(t)=-t^2+2t-\dfrac{3}{4}.

Find: The value of (a2+b2)(a^2+b^2).

First, find the maximum value of the quadratic function.

f(t)=−t2+2t−34f(t)=-t^2+2t-\frac{3}{4}

For a quadratic At2+Bt+CAt^2+Bt+C opening downward, the maximum occurs at

t=−B2A=1t=\frac{-B}{2A}=1

Hence,

fmax⁡=−1+2−34=14f_{\max}=-1+2-\frac{3}{4}=\frac{1}{4}

Therefore, the eccentricity is

e=14e=\frac{1}{4}

Now use the latus rectum formula for the ellipse:

Length of latus rectum=2b2a=30\text{Length of latus rectum}=\frac{2b^2}{a}=30

So,

b2=15ab^2=15a

Using the eccentricity relation for an ellipse,

e2=1−b2a2e^2=1-\frac{b^2}{a^2}

Substitute e=14e=\frac{1}{4} and b2=15ab^2=15a:

116=1−15aa2\frac{1}{16}=1-\frac{15a}{a^2} 116=1−15a\frac{1}{16}=1-\frac{15}{a} 1516=15a\frac{15}{16}=\frac{15}{a}

Thus,

a=16a=16

Now compute b2b^2:

b2=15a=15×16=240b^2=15a=15\times 16=240

Therefore,

a2+b2=162+240=256+240=496a^2+b^2=16^2+240=256+240=496

So, the correct option is D.

Using vertex form of the quadratic

Given: f(t)=−t2+2t−34f(t)=-t^2+2t-\dfrac{3}{4} and the eccentricity equals its maximum value.

Find: (a2+b2)(a^2+b^2) for the ellipse.

Rewrite the function in vertex form:

f(t)=−(t2−2t)−34=−[(t−1)2−1]−34=−(t−1)2+14\begin{aligned} f(t)&=-\left(t^2-2t\right)-\frac{3}{4}\\ &=-\left[(t-1)^2-1\right]-\frac{3}{4}\\ &=-(t-1)^2+\frac{1}{4} \end{aligned}

Since −(t−1)2≤0-(t-1)^2\le 0, the maximum value is 14\frac{1}{4}. Hence,

e=14e=\frac{1}{4}

For the ellipse,

2b2a=30⇒b2=15a\frac{2b^2}{a}=30 \Rightarrow b^2=15a

and

e2=1−b2a2e^2=1-\frac{b^2}{a^2}

Substituting e=14e=\frac{1}{4} gives

116=1−b2a2\frac{1}{16}=1-\frac{b^2}{a^2}

Now replace b2b^2 by 15a15a:

116=1−15aa2=1−15a\frac{1}{16}=1-\frac{15a}{a^2}=1-\frac{15}{a}

So,

15a=1516\frac{15}{a}=\frac{15}{16}

which gives

a=16a=16

Then,

b2=15⋅16=240b^2=15\cdot 16=240

Hence,

a2+b2=256+240=496a^2+b^2=256+240=496

Therefore, the answer is 496496.

Common mistakes

  • Taking the maximum value of f(t)f(t) incorrectly. A common error is to substitute a wrong vertex formula or sign for the quadratic coefficient. Since the coefficient of t2t^2 is negative, the parabola opens downward and the vertex gives the maximum value.

  • Using the wrong latus rectum formula. For the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 with major axis along the xx-axis, the length of the latus rectum is 2b2a\dfrac{2b^2}{a}, not 2a2b\dfrac{2a^2}{b}.

  • Confusing b2b^2 with bb after getting 2b2a=30\dfrac{2b^2}{a}=30. The correct rearrangement is b2=15ab^2=15a. Replacing it by b=15ab=15a leads to an incorrect eccentricity equation.

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