Let x^2f(a^2+7a+3)+y^2f(3a+15)=1 represent an ellipse with major axis along y-axis, where …

Question

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 $R$. If the set of all possible values of $a$ is $R-[\alpha ,\beta ]$, then ${\alpha }^{2}+{\beta }^{2}$ is equal to :

Accepted Answer

Since the ellipse has its major axis along the $y$-axis, the denominator under ${y}^{2}$ must be bigger than the denominator under ${x}^{2}$.

So we need

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

Given that $f$ is a strictly decreasing positive function, a bigger function value happens when the input is smaller. So,

$3a+15Bring all terms to one side:

${a}\^{2}+7a+3-3a-15>0$

${a}\^{2}+4a-12>0.$

Complete the square:

${a}\^{2}+4a-12=(a+2{)}\^{2}-16.$

So the inequality becomes

$(a+2{)}\^{2}-16>0\, \Rightarrow \, (a+2{)}\^{2}>16.$

Taking square roots,

$|a+2|>4\, \Rightarrow \, a+2>4a+2So,

$a>2aThat means the allowed values are

$a\in R-[-6,2].$

Hence, $\alpha =-6$ and $\beta =2$. Therefore,

$

Solutions · 1