Question

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

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

Answer 1

The correct option is B $\frac{1}{a}$

Distance formula: The distance between points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$A=({at}^2,2at),B=(\frac{a}{t^2},\frac{-2a}{t}),S(a,0)SA=\sqrt{{(a-{at}^2)}^2+{(0-2at)}^2}=\sqrt{a^2+a^2{t}^4-2a^2{t}^2+4a^2{t}^2}=\sqrt{a^2+a^2{t}^4+2a^2{t}^2}=\sqrt{{(a+{at}^2)}^2}=a+{at}^{2}SB=\sqrt{{(a-\frac{a}{t^2})}^2+{(0+\frac{2a}{t})}^2}=\sqrt{a^2+\frac{a^2}{t^4}+\frac{4a^2}{t^2}-\frac{2a^2}{t^2}}=\sqrt{a^2+\frac{a^2}{t^4}+\frac{2a^2}{t^2}}=\sqrt{{(a+\frac{a}{t^2})}^2}=a+\frac{a}{t^2}Now,\frac{1}{SA}+\frac{1}{SB}=\frac{1}{a+{at}^2}+\frac{1}{a+\frac{a}{t^2}}=\frac{1}{a+{at}^2}+\frac{t^2}{a+{at}^2}=\frac{(1+t^2)}{a(1+t^2)}=\frac{1}{a}$