Question
Let $\frac{x^{2}}{f\left( a^{2} + 7a + 3 \right)} + \frac{y^{2}}{f(3a + 15)} = 1$ 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 $\mathbf{R} - \lbrack\alpha,\beta\rbrack$, then $\alpha^{2} + \beta^{2}$ is equal to :
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Solution
Sol. [Key Concept: The major axis on the $y$-axis needs $f(3a + 15) > f(a^{2} + 7a + 3)$; since $f$ decreases this gives $a^{2} + 4a - 12 > 0$, so $a \in mathbbR - \lbrack - 6,2\rbrack$ and $\alpha^{2} + \beta^{2} = 40$.]
Given ellipse is vertical
$$\therefore f(3a + 15) > f\left( a^{2} + 7a + 3 \right) $$$\because f(x)$ is decreasing $\forall x \in R$
$${\Rightarrow 3a + 15 < a^{2} + 7a + 3 }{\Rightarrow a^{2} + 4a - 12 > 0 }{\Rightarrow (a + 2)^{2} > 16 }{\Rightarrow |(a + 2)| > 4 }$$$\Rightarrow a > 2$ or $a < - 6$
$${\Rightarrow a \in R - \lbrack - 6,2\rbrack }{\therefore\alpha = - 6,\beta = 2 }{\therefore\alpha^{2} + \beta^{2} = 36 + 4 = 40}$$
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