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(D) The length of the focal chord of a parabola $y^2=4ax$ is given by $L = 4a \operatorname{cosec}^2 	heta$,where $	heta$ is the angle the chord mak

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$72$

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(D) The length of the focal chord of a parabola $y^2=4ax$ is given by $L = 4a \operatorname{cosec}^2 	heta$,where $	heta$ is the angle the chord makes with the axis of the parabola.Here,$4a = 12$,so $a = 3$. The length $L = 15$.$12 \operatorname{cosec}^2 	heta = 15 \implies \operatorname{cosec}^2 	heta = \frac{15}{12} = \frac{5}{4}$.Then $\sin^2 	heta = \frac{4}{5}$,which implies $\cos^2 	heta = 1 - \frac{4}{5} = \frac{1}{5}$.Thus,$	an^2 	heta = \frac{\sin^2 	heta}{\cos^2 	heta} = \frac{4/5}{1/5} = 4$,so $	an 	heta = 2$.The focal chord passes through the focus $(a, 0) = (3, 0)$. The equation of the chord with slope $m = 	an 	heta$ is $y - 0 = m(x - 3)$.Since the chord makes an angle $	heta$ with the axis,its slope is $m = \pm 	an 	heta = \pm 2$. Let $m = 2$.The equation is $y = 2(x - 3) \implies 2x - y - 6 = 0$.The distance $p$ of the line $2x - y - 6 = 0$ from the origin $(0, 0)$ is $p = \frac{|2(0) - 0 - 6|}{\sqrt{2^2 + (-1)^2}} = \frac{6}{\sqrt{5}}$.Therefore,$p^2 = \frac{36}{5}$.Finally,$10p^2 = 10 	imes \frac{36}{5} = 2 	imes 36 = 72$.

Let the length of the focal chord $PQ$ of the parabola $y^2=12x$ be $15$ units. If the distance of $PQ$ from the origin is $p$,then $10p^2$ is equal to:

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