Maths Q12 – JEE-MAIN 2025

Let the length of a latus rectum of an ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

be 10. If its eccentricity is the minimum value of the function

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

t \in R

, then

a^{2} + b^{2}

is equal to :

Clue 1 of 3: Understanding Eccentricity

The eccentricity of an ellipse is a measure of its deviation from being circular. It is given as the minimum value of a quadratic function.

🧠 Full Solution Path

Step 1: Determine the eccentricity

e

✦ Active

The eccentricity $e$ is the minimum value of the function $f (t) = t^{2} + t + \frac{11}{12}$ . To find the minimum value, we complete the square:

f (t) = (t^{2} + t + \frac{1}{4}) - \frac{1}{4} + \frac{11}{12}

f (t) = (t + \frac{1}{2})^{2} - \frac{3}{12} + \frac{11}{12}

f (t) = (t + \frac{1}{2})^{2} + \frac{8}{12}

f (t) = (t + \frac{1}{2})^{2} + \frac{2}{3}

The minimum value of $f (t)$ is $\frac{2}{3}$ , which occurs when $t = - \frac{1}{2}$ . Therefore, the eccentricity $e = \frac{2}{3}$ .

Step 2: Use the latus rectum length to find a relation between

a

and

b

○ Expand

The length of the latus rectum of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is given as 10. The formula for the length of the latus rectum is $\frac{2 b^{2}}{a}$ .

\frac{2 b^{2}}{a} = 10 ⟹ b^{2} = 5 a

Step 3: Calculate

a

and

b^{2}

using the eccentricity relation○ Expand

For an ellipse, the relationship between $a, b,$ and $e$ is $b^{2} = a^{2} (1 - e^{2})$ . Substitute $b^{2} = 5 a$ and $e = \frac{2}{3}$ into this equation:

5 a = a^{2} (1 - (\frac{2}{3})^{2})

5 a = a^{2} (1 - \frac{4}{9})

5 a = a^{2} (\frac{5}{9})

Since $a$ is the semi-major axis, $a \neq 0$ . We can divide by $a$ :

5 = a (\frac{5}{9}) ⟹ a = 9

Now, find $b^{2}$ using $b^{2} = 5 a$ :

b^{2} = 5 (9) = 45

Step 4: Compute

a^{2} + b^{2}

○ Expand

Finally, calculate $a^{2} + b^{2}$ :

a^{2} = 9^{2} = 81

a^{2} + b^{2} = 81 + 45 = 126

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Maths Question 12 – JEE-MAIN 2025

Clue 1 of 3: Understanding Eccentricity

🧠 Full Solution Path