Question

If $P$ and $Q$ are two points whose coordinates are $(a{t}^2,2at)$ and $(\frac{a}{t^2},-\frac{2a}{t})$ respectively and $S$ is the point $(a,0)$. Show that $\frac{1}{SP}+\frac{1}{SQ}$ is independent of $t$.

Answer 1

$P(a{t}^2,2at)$ $Q(\frac{a}{t^2},\frac{-2a}{t})$ $S(a,0)$

$SP=\sqrt{(a{t}^2-a{)}^2+(2at{)}^2}$

$=\sqrt{a^2(t^2-1{)}^2+a^{2}4t^2}=a\sqrt{t^4+1+2t^2}$

$=a\sqrt{(t^2+1{)}^2}=a(t^2+1)$

$SQ=\sqrt{{(\frac{a}{t^2}-a)}^2+(\frac{2a}{t}{)}^2}$

$=\sqrt{a^2{(\frac{1}{t^2}-1)}^2+{\left(\frac{2}{t}\right)}^2{a}^2}=a\sqrt{\frac{1}{t^4}+1+\frac{2}{t^2}}$

$=a\sqrt{{(\frac{1}{t^2}+1)}^2}=a(\frac{1}{t^2}+1)$

$\frac{1}{SP}+\frac{1}{SQ}=\frac{1}{a(t^2+1)}+\frac{1}{a(\frac{1}{t^2}+1)}$

$=\frac{1}{a(t^2+1)}+\frac{t^2}{a(1+t^2)}=\frac{1+t^2}{a(t^2+1)}=\frac{1}{a}$

as you can see it is independent of $t$

Hence Proved