Question

If $A(a^2,2a)B=(\frac{1}{a^2},\frac{-2}{a}),P=(1,0)$ then $\frac{1}{PA}+\frac{1}{PB}=$

Answer 1

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

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

$\frac{1}{PA}+\frac{1}{PB}=\frac{1}{1+a^2}+\frac{a^2}{1+a^2}=1$