Let a parabola P be such that its vertex and | vedclass

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(C) The vertex of the parabola is $V(2,0)$ and the focus is $F(4,0)$.Thus,the distance $VF = a = 4 - 2 = 2$.The equation of the parabola is $(y - 0)^2

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$16$

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(C) The vertex of the parabola is $V(2,0)$ and the focus is $F(4,0)$.Thus,the distance $VF = a = 4 - 2 = 2$.The equation of the parabola is $(y - 0)^2 = 4a(x - 2)$,which simplifies to $y^2 = 8(x - 2)$.Let the tangent from the origin $O(0,0)$ to the parabola be $y = mx + c$. Since it passes through $(0,0)$,$c = 0$,so $y = mx$.Substituting $y = mx$ into $y^2 = 8x - 16$,we get $(mx)^2 = 8x - 16$,or $m^2x^2 - 8x + 16 = 0$.For tangency,the discriminant $D = (-8)^2 - 4(m^2)(16) = 0$.$64 - 64m^2 = 0$ $\Rightarrow m^2 = 1$ $\Rightarrow m = \pm 1$.The points of contact $S$ and $R$ are found by solving $x^2 - 8x + 16 = 0$ (for $m=1$) and $(-x)^2 - 8x + 16 = 0$ (for $m=-1$).Both give $(x-4)^2 = 0$,so $x = 4$.For $x = 4$,$y = \pm 4$. Thus,the points are $R(4, 4)$ and $S(4, -4)$.The base of $	riangle SOR$ is the segment $RS$,which has length $4 - (-4) = 8$.The height of $	riangle SOR$ with respect to base $RS$ is the distance from $O(0,0)$ to the line $x = 4$,which is $4$.Area of $	riangle SOR = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 8 	imes 4 = 16 	ext{ sq. units}$.

Let a parabola $P$ be such that its vertex and focus lie on the positive $x$-axis at a distance $2$ and $4$ units from the origin,respectively. If tangents are drawn from $O(0,0)$ to the parabola $P$ which meet $P$ at $S$ and $R$,then the area (in $sq. \text{ units}$) of $\triangle SOR$ is equal to:

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