Question

If $A(a^2,2a)B(\frac{1}{a^2},\frac{-2}{a})$ and S(1,0) then prove that $\frac{1}{SA}+\frac{1}{SB}=1$

Answer 1

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

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

$=\sqrt{1+a^4+2a^2}$

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

$=(1+a^2)$ ..........1

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

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

$=\sqrt{1+\frac{1}{a^4}+\frac{2}{a^2}}$

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

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

$=\frac{a^2+1}{a^2}$ ......2

Then,$\frac{1}{SA}+\frac{1}{SB}=\frac{1}{1+a^2}+\frac{a^2}{1+a^2}$

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