Question

Let $f(x)=\sqrt{\left(x-\frac{15}{4}\right)^2+2-\frac{x^2}{16}}+\sqrt{(x-6)^2+\left(\sqrt{2-\frac{x^2}{16}}-9\right)^2}$ where $-4 \sqrt{2} \leq x \leq 4 \sqrt{2}$ If the least value of $f(x)$ is $\frac{p}{q} \sqrt{17}$, where $p, q$ are relatively prime numbers, then find the value of $(p-q)$

Answer 1

$f(x)=\sqrt{\left(x-\frac{15}{4}\right)^2+2-\frac{x^2}{16}}+\sqrt{\left(x-6^2\right)+\left(\sqrt{2-\frac{x^2}{16}}-9\right)^2}$

$P \equiv\left( x , \sqrt{2-\frac{ x ^2}{16}}\right), A \equiv\left(\frac{15}{4}, 0\right), B \equiv(6,9)$
P lies on ellipse $\frac{x^2}{16}+\frac{y^2}{1}=2$
$PA + PB \geq AB$
$\therefore $ Minimum value of $PA + PB = AB$.
$AB =\sqrt{\left(6-\frac{15}{4}\right)^2+(9-0)^2}=\sqrt{\frac{81}{16}+81}=\frac{9}{4} \sqrt{17} $
$\therefore p - q =5 $