JEE Main 2026 (Online) 8th April Evening ShiftQuestion 1
Let $\frac{x^{2}}{f(a^{2}+7a+3)}+\frac{y^{2}}{f(3a+15)}=1$ represent an ellipse with major axis along $y$-axis, where $f$ is a strictly decreasing positive function on $\mathbf{R}$. If the set of all possible values of $a$ is $\mathbf{R}-[\alpha,\beta]$, then $\alpha^{2}+\beta^{2}$ is equal to :
Ellipse
Solution
Concept. For $\frac{x^2}{A}+\frac{y^2}{B}=1$ (both positive) the major axis is along $y$ exactly when $B>A$. A strictly decreasing $f$ reverses inequalities. Formulas used. (1) Major axis along $y$ $\iff$ denominator of $y^2$ > denominator of $x^2$(2) $f$ strictly decreasing: $f(u)>f(v)\iff u<v$Step 1 - Condition from (1). Both denominators are positive since $f>0$. Major axis along $y$ needs $$f(3a+15)>f(a^2+7a+3).$$ Step 2 - Use (2). Since $f$ is strictly decreasing, this means $$3a+15<a^2+7a+3.$$ Step 3 - Solve. $a^2+4a-12>0\Rightarrow(a+6)(a-2)>0$, so $a<-6$ or $a>2$, i.e. $a\in\mathbb R-[-6,2]$. Step 4 - Compute. $\alpha=-6,\ \beta=2$, so $\alpha^2+\beta^2=36+4=40$. Answer: B