Let … represent an ellipse with major…

Single correct+4 / -1Answer matches NTA’s final key

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

Solution

Major axis along $y$ means the $y^2$ denominator is larger:

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

$f$ is strictly decreasing, so this reverses on the arguments:

$3a + 15 < a^2 + 7a + 3 \iff a^2 + 4a - 12 > 0 \iff (a + 6)(a - 2) > 0.$

So $a \in \mathbb{R} - [-6, 2]$ , giving $\alpha = -6$ , $\beta = 2$ and $\alpha^2 + \beta^2 = 36 + 4 = 40$ .

Answer: 40.

Answer checked against the official answer key. The solution is our own working.

JEE MAIN 2026 · Co-ordinate Geometry

Solution

Keep going