Mathematics · Ellipse

JEE Main 2026 — 8 April, Evening Shift — Question 37

Let $\frac{x^{2}}{f(a^{2}+7a+3)} + \frac{y^{2}}{f(3a+15)} = 1$ represent an ellipse with major axis along y-axis, where f is a strictly decreasing positive function on R. If the set of all possible values of a is $R - [α,β],$ then $α²+β²$ is equal to:

Option A:
$28$
Option B:
$40$
Correct
Option C:
$61$
Option D:
$24$

Answer: B

Step-by-step solution

Given ellipse is vertical $\therefore \mathrm{f}(3 \mathrm{a}+15)>\mathrm{f}\left(\mathrm{a}^{2}+7 \mathrm{a}+3\right)$ $\because \mathrm{f}(\mathrm{x})$ is decreasing $\forall \mathrm{x} \in \mathrm{R}$ $\Rightarrow 3 \mathrm{a}+15<\mathrm{a}^{2}+7 \mathrm{a}+3$ $\Rightarrow \mathrm{a}^{2}+4 \mathrm{a}-12>0$ $\Rightarrow(\mathrm{a}+2)^{2}>16$ $\Rightarrow|(\mathrm{a}+2)|>4$ $\Rightarrow \mathrm{a}>2$ or $\mathrm{a}<-6$ $\Rightarrow \mathrm{a} \in \mathrm{R}-[-6,2]$ $\therefore \alpha=-6, \beta=2$ $\therefore \alpha^{2}+\beta^{2}=36+4=40$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 8 April, Evening Shift
Subject: Mathematics
Chapter: Ellipse
Topic: Introduction to Ellipse