JEE Main 2026 — Ellipse Question with Solution

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

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

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

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

$a^2+4a-12>0$

$(a+6)(a-2)>0$

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

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

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

${\alpha }^2+{\beta }^2=(-6{)}^2+(2{)}^2=36+4=40$

Answer: $40$