P and Q are the extremities of a focal chord | vedclass

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(C) Given the parabola $y^2=4ax$. Since $P(9,9)$ lies on the parabola,we have $81=4 	imes a 	imes 9$,which gives $a=\frac{9}{4}$.We know that the ex

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$\frac{45}{16}$

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(C) Given the parabola $y^2=4ax$. Since $P(9,9)$ lies on the parabola,we have $81=4 	imes a 	imes 9$,which gives $a=\frac{9}{4}$.We know that the extremities of a focal chord are given by $P(at^2, 2at)$ and $Q(\frac{a}{t^2}, -\frac{2a}{t})$.Comparing $P(9,9)$ with $(at^2, 2at)$,we get $2at=9$ $\Rightarrow 2(\frac{9}{4})t=9$ $\Rightarrow t=2$.Thus,$Q = (\frac{a}{t^2}, -\frac{2a}{t}) = (\frac{9/4}{4}, -\frac{2(9/4)}{2}) = (\frac{9}{16}, -\frac{9}{4})$.Therefore,$p=\frac{9}{16}$ and $q=-\frac{9}{4}$.Then $p-q = \frac{9}{16} - (-\frac{9}{4}) = \frac{9}{16} + \frac{36}{16} = \frac{45}{16}$.

$P$ and $Q$ are the extremities of a focal chord of the parabola $y^2=4ax$. If $P=(9,9)$ and $Q=(p, q)$,then $p-q=$

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