Question

A focal chord of the parabola $y^2=4ax$ meets it at P and Q. If S is the focus then $\frac{1}{SP}+\frac{1}{SQ}=$

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

Answer 1

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

Let $P=(a{t}_1^2,2a{t}_1),Q=(a{t}_2^2,2a{t}_2)$ Now focus S=(a,0)
Since PQ is a focal chord,$t_1{t}_2=-1$
$SP=\sqrt{(a{t}_1^2-a{)}^2+(2a{t}_1-0{)}^2}=a\sqrt{(t_1^2-1{)}^2+4t_1^2}=a\sqrt{(t_1^2-1{)}^2}=a(t_1^2-1)$
$SQ=\sqrt{(a{t}_2^2-a{)}^2+(2a{t}_2-0{)}^2}=a\sqrt{(t_2^2-1{)}^2+4t_1^2}=a\sqrt{(t_2^2-1{)}^2}=a(t_2^2-1)$
$\frac{1}{SP}+\frac{1}{SQ}=\frac{1}{a(t_1^2+1)}+\frac{1}{a(t_2^2+1)}=\frac{1}{a}[\frac{1}{t_1^2+1}+\frac{1}{t_2^2+1}]=\frac{1}{a}[\frac{t_2^2+1+t_2^2+1}{(t_2^2+1)(t_2^2+1)}+]=\frac{1}{a}\left(\frac{t_1^2+t_2^2+2}{t_1^2+t_1^2{t}_2^2+t_2^2+1}\right)$
$=\frac{1}{a}\left[\frac{t_1^2+t_2^2+2}{1+t_1^2+t_2^2+1}\right]=\frac{1}{a}$