JEE Main 2026 — Mathematics Coordinate Geometry

For the given equation to represent an ellipse with its major axis along the $y$y-axis, the denominator of $y^2$y2 must be strictly greater than the denominator of $x^2$x2.

$$f(3a+15) > f(a^2+7a+3)$$f(3a+15)>f(a2+7a+3)

Since $f$f is given as a strictly decreasing positive function on $\mathbb{R}$R, the inequality sign reverses for the arguments:

$$3a+15 < a^2+7a+3$$3a+15<a2+7a+3

$$a^2 + 4a - 12 > 0$$a2+4a−12>0

$$(a+6)(a-2) > 0$$(a+6)(a−2)>0

The solution to this inequality is $a \in (-\infty, -6) \cup (2, \infty)$a∈(−∞,−6)∪(2,∞).

This can be rewritten in terms of the set difference as $a \in \mathbb{R} - [-6, 2]$a∈R−[−6,2].

Comparing this with the given set $\mathbb{R} - [\alpha, \beta]$R−[α,β], we get $\alpha = -6$α=−6 and $\beta = 2$β=2.

$$\alpha^2 + \beta^2 = (-6)^2 + (2)^2 = 36 + 4 = 40$$α2+β2=(−6)2+(2)2=36+4=40

Answer: $40$40