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For the parabola $\mathrm{y^2=4 a x, A B}$ is a focal chord with $\mathrm{A\left(a t_1^2, 2 a t_1\right)}$ and $\mathrm{B\left(a t_2^2, 2 a t_2\right)}$ as its end points; $\mathrm{S}$ is the focus of parabola with coordinates $\mathrm{(a, 0)}$ . Then, $\mathrm{S A, 2 a}$ and $\mathrm{S B}$ are in

Option: 1
A.P.

Option: 2
G.P.

Option: 3
H.P.

Option: 4
none of these

Answers (1)

best_answer

As $\mathrm{S(a, 0)}$ is the focus of the parabola and the directrix is $\mathrm{x+a=0,}$

$\mathrm{S A =A M}$ (by definition of parabola)

$\mathrm{ =a+a t_1^2 }$

$\mathrm{S B =B N }$

$\mathrm{ =a+a t_1^2 }$
$\mathrm{ \left.=a+a\left(\frac{-1}{t_1}\right)^2 \text { (as for focal chord } A B, t_1 t_2=-1\right) }$
$\mathrm{ =\frac{a t_1^2+a}{t_1^2} }$

$\mathrm{ \therefore \frac{1}{S A}+\frac{1}{S B}=\frac{1+t_1^2}{a\left(1+t_1^2\right)}=\frac{2}{2 a} }$

$\mathrm{ \Rightarrow S A, 2 a}$ (semi latus rectum), SB are in H.P.

Hence option 3 is correct.

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