Let P represent the point (3, 6) on the parab | vedclass

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(A) The equation of the parabola is $y^2 = 12x$,which is of the form $y^2 = 4ax$,so $a = 3$. Point $P(3, 6)$ corresponds to $at^2 = 3$ and $2at = 6$.

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$2 \sqrt{2}$

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(A) The equation of the parabola is $y^2 = 12x$,which is of the form $y^2 = 4ax$,so $a = 3$. Point $P(3, 6)$ corresponds to $at^2 = 3$ and $2at = 6$. Substituting $a = 3$,we get $3t^2 = 3 \implies t = 1$. The length of the normal chord at $t$ is given by $l_1 = 2a(t^2+2) \sqrt{t^2+1}$. For $t = 1$,$l_1 = 2(3)(1+2) \sqrt{1+1} = 6(3) \sqrt{2} = 18 \sqrt{2}$. The length of the focal chord passing through $P(at^2, 2at)$ is $l_2 = a(t + \frac{1}{t})^2$. For $t = 1$,$l_2 = 3(1 + \frac{1}{1})^2 = 3(2)^2 = 12$. Therefore,$\frac{l_1}{l_2} = \frac{18 \sqrt{2}}{12} = \frac{3 \sqrt{2}}{2} = \frac{3}{\sqrt{2}}$. Wait,re-evaluating the normal chord length formula: The length of the normal chord at $t$ is $l_1 = 2a(t^2+2) \sqrt{1+t^2}$. For $t=1$,$l_1 = 2(3)(3)\sqrt{2} = 18\sqrt{2}$. Re-evaluating the focal chord length: The length of a focal chord with parameter $t$ is $a(t + 1/t)^2$. For $t=1$,$l_2 = 3(1+1)^2 = 12$. Thus,$\frac{l_1}{l_2} = \frac{18\sqrt{2}}{12} = \frac{3\sqrt{2}}{2}$.

Let $P$ represent the point $(3, 6)$ on the parabola $y^2 = 12x$. For the parabola $y^2 = 12x$,if $l_1$ is the length of the normal chord drawn at $P$ and $l_2$ is the length of the focal chord drawn through $P$,then $\frac{l_1}{l_2} = $

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