JEE Main 2026 Apr 8 Shift 2, Mathematics Q20: Introduction to Ellipse

JEE Main2026Apr 8, Shift 2Mathematics

Let $\frac{x ^{2}}{f ( a ^{2} + 7 a + 3 )} + \frac{y ^{2}}{f ( 3 a + 15 )} = 1$ 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 $α^{2} + β^{2}$ is equal to :

A
$28$
B
$40$
C
$61$
D
$24$

Solution

Major axis along $y$ -axis means the denominator of $y^{2}$ exceeds the denominator of $x^{2}$ :

$f (3 a + 15) > f (a^{2} + 7 a + 3)$

Since $f$ is strictly decreasing, this inverts to:

$3 a + 15 < a^{2} + 7 a + 3 ⟹ a^{2} + 4 a - 12 > 0 ⟹ (a + 6) (a - 2) > 0$

So $a < - 6$ or $a > 2$ , i.e. $a \in R - [- 6, 2]$ .

Therefore $α = - 6$ , $β = 2$ , and $α^{2} + β^{2} = 36 + 4 = 40$ .

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JEE Main 2026 Apr 8 Shift 2, Mathematics Q20: Introduction to Ellipse

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