Consider the parabola y^2=8x. Let Delta_1 be | vedclass

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(B) The given equation of the parabola is $y^2=8x$,which is of the form $y^2=4ax$. Thus,$4a=8$,so $a=2$.The endpoints of the latus rectum are $(a, 2a)

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$2$

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(B) The given equation of the parabola is $y^2=8x$,which is of the form $y^2=4ax$. Thus,$4a=8$,so $a=2$.The endpoints of the latus rectum are $(a, 2a)$ and $(a, -2a)$,which are $(2, 4)$ and $(2, -4)$.The area $\Delta_1$ of the triangle formed by $(2, 4)$,$(2, -4)$,and $P\left(\frac{1}{2}, 2\right)$ is given by the determinant formula:$\Delta_1 = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$$\Delta_1 = \frac{1}{2} |2(-4-2) + 2(2-4) + 0.5(4-(-4))| = \frac{1}{2} |-12 - 4 + 4| = \frac{1}{2} |-12| = 6$.The tangent at $(x_1, y_1)$ to $y^2=8x$ is $yy_1 = 4(x+x_1)$.Tangent at $(2, 4)$: $4y = 4(x+2) \Rightarrow y = x+2$.Tangent at $(2, -4)$: $-4y = 4(x+2) \Rightarrow y = -x-2$.Tangent at $(0.5, 2)$: $2y = 4(x+0.5) \Rightarrow y = 2x+1$.Intersection points:$1$. Tangents at $(2, 4)$ and $(2, -4)$: $x+2 = -x-2$ $\Rightarrow 2x = -4$ $\Rightarrow x = -2, y = 0$. Point: $(-2, 0)$.$2$. Tangents at $(2, 4)$ and $(0.5, 2)$: $x+2 = 2x+1 \Rightarrow x = 1, y = 3$. Point: $(1, 3)$.$3$. Tangents at $(2, -4)$ and $(0.5, 2)$: $-x-2 = 2x+1$ $\Rightarrow 3x = -3$ $\Rightarrow x = -1, y = -1$. Point: $(-1, -1)$.Area $\Delta_2 = \frac{1}{2} |(-2)(3 - (-1)) + 1(-1 - 0) + (-1)(0 - 3)| = \frac{1}{2} |-8 - 1 + 3| = \frac{1}{2} |-6| = 3$.Therefore,$\frac{\Delta_1}{\Delta_2} = \frac{6}{3} = 2$.

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