If the normal drawn at the point P(9, 9) on t | vedclass

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(D) The equation of the parabola is $y^2 = 9x$. Comparing with $y^2 = 4ax$,we get $4a = 9$,so $a = \frac{9}{4}$.At point $P(9, 9)$,let the parameter b

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$27$

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(D) The equation of the parabola is $y^2 = 9x$. Comparing with $y^2 = 4ax$,we get $4a = 9$,so $a = \frac{9}{4}$.At point $P(9, 9)$,let the parameter be $t_1$. Then $2at_1 = 9$ $\Rightarrow 2(\frac{9}{4})t_1 = 9$ $\Rightarrow t_1 = 2$.The normal at $t_1$ meets the parabola at $t_2$,where $t_2 = -t_1 - \frac{2}{t_1} = -2 - \frac{2}{2} = -3$.The coordinates of $Q(a, b)$ are $(at_2^2, 2at_2) = (\frac{9}{4} 	imes (-3)^2, 2 	imes \frac{9}{4} 	imes (-3)) = (\frac{81}{4}, -\frac{27}{2})$.Thus,$a = \frac{81}{4}$ and $b = -\frac{27}{2}$.Calculating $2a + b = 2(\frac{81}{4}) + (-\frac{27}{2}) = \frac{81}{2} - \frac{27}{2} = \frac{54}{2} = 27$.

If the normal drawn at the point $P(9, 9)$ on the parabola $y^2 = 9x$ meets the parabola again at $Q(a, b)$,then $2a + b =$

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