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 $-4\sqrt{2}\le x\le 4\sqrt{2}$ 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

Answer: 5

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

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