Question

$A\equiv (t^2,2t),B\equiv \left(\frac{1}{t^2},-\frac{2}{t}\right)ands\equiv (1,0),then\frac{1}{SA}+\frac{1}{SB}isequalto$

• A. $2$
• B. $\frac{1}{2}$
• C. $\frac{1}{5}$
• D. $1$

Answer 1

The correct option is D $1$

$A=(t^2,2t)$

$B=(\frac{1}{t^2},\frac{-2}{t})$

$S=(1,0)$

Distance formula$=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$SA=\sqrt{(t^2-1{)}^2+(2t-0{)}^2}$

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

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

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

$SA=t^2+1$

$SB=\sqrt{{(\frac{1}{t^2}-1)}^2+{(-\frac{2}{t}-0)}^2}$

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

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

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

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

$SB=\frac{1+t^2}{t^2}$

Now $\frac{1}{SA}+\frac{1}{SB}$

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