JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Ellipse

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

R - [α, β]

, then

α^{2} + β^{2}

is equal to :

Answer: (B)

Solution

The standard-form ellipse has its major axis along the

y

-axis when the denominator below

y^{2}

is the larger, so

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

. A strictly decreasing function reverses the order of its arguments:

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

giving

a < - 6

a > 2

, that is

a \in R - [- 6, 2]

. With

α = - 6

and

β = 2

α^{2} + β^{2} = 36 + 4 = 40.

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option