5 ; $\sqrt{3^2+4^2}=5$

$y$ ; $\sqrt{x^2+{\left(y-2\right)}^2}$

$y^2=x^2+{\left(y-2\right)}^2$
$y^2=x^2+y^2-4y+4$
$0=x^2-4y+4$
$4y=x^2+4$
$y=\frac{1}{4}{x}^2+1$

$3=\frac{1}{4}{x}^2+1$
$2=\frac{1}{4}{x}^2$
$8=x^2$
$x=\sqrt{8}\approx \mathrm{2,83}$$$of$$$x=\mathrm{‐}\sqrt{8}\approx \mathrm{‐}\mathrm{2,83}$

$4=\frac{1}{4}{x}^2+1$
$3=\frac{1}{4}{x}^2$
$12=x^2$
$x=\sqrt{12}\approx \mathrm{3,46}$$$of$$$x=\mathrm{‐}\sqrt{12}\approx \mathrm{‐}\mathrm{3,46}$

$6=\frac{1}{4}{x}^2+1$
$5=\frac{1}{4}{x}^2$
$20=x^2$
$x=\sqrt{20}\approx \mathrm{4,47}$$$of$$$x=\mathrm{‐}\sqrt{20}\approx \mathrm{‐}\mathrm{4,47}$

$P$: Afstand tot $E$ is $\sqrt{{\left(1\frac{1}{2}\right)}^2+{\left(1\frac{3}{4}\right)}^2}$ = $\sqrt{\frac{85}{16}}$ = $\frac{1}{4}\sqrt{85}$
Afstand tot $k$ is $2\frac{1}{4}$
$Q$: Afstand tot $E$ is $\sqrt{{\left(\mathrm{‐3}\right)}^2+{\left(8\frac{3}{4}\right)}^2}$ = $\sqrt{\frac{1369}{16}}$ = $\frac{37}{4}$ = $9\frac{1}{4}$
Afstand tot $k$ is $9\frac{1}{4}$
$R$: Afstand tot $E$ is $\sqrt{8^2+{\left(63\frac{3}{4}\right)}^2}$ = $\sqrt{\frac{66.049}{16}}$ = $\frac{257}{4}$ = $64\frac{1}{4}$
Afstand tot $k$ is $64\frac{1}{4}$
Dus $Q$ en $R$ liggen even ver van $E$ als van $k$.

De afstand tot $k$ is $y+\frac{1}{4}$
De afstand tot $E$ is $\sqrt{x^2+{\left(y-\frac{1}{4}\right)}^2}$
Dus
$y+\frac{1}{4}$ = $\sqrt{x^2+{\left(y-\frac{1}{4}\right)}^2}$ $\Rightarrow$
${\left(y+\frac{1}{4}\right)}^2=x^2+{\left(y-\frac{1}{4}\right)}^2$

${\left(y+\frac{1}{4}\right)}^2=x^2+{\left(y-\frac{1}{4}\right)}^2$
$y^2+\frac{1}{2}y+\frac{1}{16}=x^2+y^2-\frac{1}{2}y+\frac{1}{16}$
$y=x^2$