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(A) Let the tangent to the parabola $y^2=4ax$ at point $P(at^2, 2at)$ be $yt = x + at^2$ ... $(i)$. Since the line $NM$ is perpendicular to the tangen

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$4a^2$

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(A) Let the tangent to the parabola $y^2=4ax$ at point $P(at^2, 2at)$ be $yt = x + at^2$ ... $(i)$. Since the line $NM$ is perpendicular to the tangent and passes through the origin $O(0,0)$,its equation is $y = -tx$ ... (ii). To find $N$,substitute $y = -tx$ into $(i)$: $t(-tx) = x + at^2 \implies -t^2x - x = at^2 \implies x = -\frac{at^2}{1+t^2}$. Then $y = -t(-\frac{at^2}{1+t^2}) = \frac{at^3}{1+t^2}$. So,$N \equiv (-\frac{at^2}{1+t^2}, \frac{at^3}{1+t^2})$. To find $M$,substitute $y = -tx$ into $y^2 = 4ax$: $(-tx)^2 = 4ax \implies t^2x^2 = 4ax$. Since $x 
eq 0$,$x = \frac{4a}{t^2}$. Then $y = -t(\frac{4a}{t^2}) = -\frac{4a}{t}$. So,$M \equiv (\frac{4a}{t^2}, -\frac{4a}{t})$. Now,$ON = \sqrt{(-\frac{at^2}{1+t^2})^2 + (\frac{at^3}{1+t^2})^2} = \sqrt{\frac{a^2t^4(1+t^2)}{(1+t^2)^2}} = \frac{at^2}{\sqrt{1+t^2}}$. And $OM = \sqrt{(\frac{4a}{t^2})^2 + (-\frac{4a}{t})^2} = \sqrt{\frac{16a^2}{t^4} + \frac{16a^2}{t^2}} = \frac{4a}{t^2} \sqrt{1+t^2}$. Therefore,$ON \cdot OM = \frac{at^2}{\sqrt{1+t^2}} \cdot \frac{4a}{t^2} \sqrt{1+t^2} = 4a^2$.

List $I$	List $II$
$P. \quad m=$	$1. \quad 1/2$
$Q. \quad \text{Maximum area of } \triangle EFG \text{ is}$	$2. \quad 4$
$R. \quad y_0=$	$3. \quad 2$
$S. \quad y_1=$	$4. \quad 1$

If a perpendicular drawn through the vertex $O$ of the parabola $y^2=4ax$ to any of its tangent meets the tangent at $N$ and the parabola at $M$,then $ON \cdot OM=$

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