Question

P, Q are two points whose coordinates are $(a{t}^2,2at)$ and $(\frac{a}{t^2},\frac{-2a}{t})$ and S is a point whose coordinates is $(a,0)$, then $\frac{1}{SP}+\frac{1}{SQ}$ is constant for all values of $t$.

• A. True
• B. False

Answer 1

The correct option is A True

Points, $P(a{t}^2,2at),Q(\frac{9}{t^2},\frac{-2a}{t})andS(a,0)$

$=>SP=\sqrt{a^2(t^2-{1)}^2+{\left(2at\right)}^2}=a\sqrt{(t^2-{1)}^2+4t^2}$

$=>SQ=\sqrt{{(\frac{a}{t^2}-a)}^2+{\left(\frac{2a}{t}\right)}^2}=\frac{a}{t^2}\sqrt{(1-{t^2)}^2+4t^2}$

$=>\frac{1}{SP}+\frac{1}{SQ}=\frac{1}{a}(\frac{t}{\sqrt{(1-{t^2)}^2+4t^2}}+\frac{1}{\sqrt{t^2-{1)}^2+4t^2}})$

$=\frac{1}{a}\left(\frac{t^2\sqrt{t^4+2t^2+1}+\sqrt{t^4+{2t}^2+1}}{(t^4+{2t}^2+1)}\right)$

$=\frac{1}{a}\left(\frac{(t^2+1)\left[\right(t^2+{1{)}^2]}^{\frac{1}{2}}}{{{(t}^2+1)}^2}\right)$

$==\frac{1}{a}$ $=constant$

As we know, $P,Q$ are parametric co-ordinate of parabola with focus $(Q,0)$

And $P$and$Q$ are focal chord and their distance from focus when added reciprocally gives constant $\frac{1}{a}$.