Let P(2,4) and Q(18,-12) be the points on the | vedclass

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(D) The equation of the parabola is $y^2=8x$. Comparing with $y^2=4ax$,we get $4a=8$,so $a=2$.The equation of the tangent to the parabola $y^2=4ax$ at

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$x-2y=2$

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(D) The equation of the parabola is $y^2=8x$. Comparing with $y^2=4ax$,we get $4a=8$,so $a=2$.The equation of the tangent to the parabola $y^2=4ax$ at point $(x_1, y_1)$ is $yy_1 = 2a(x+x_1)$.For point $P(2,4)$,the tangent is $4y = 4(x+2) \Rightarrow y = x+2$  $(i)$.For point $Q(18,-12)$,the tangent is $-12y = 4(x+18) \Rightarrow -3y = x+18$  $(ii)$.To find the intersection point,substitute $x = y-2$ from $(i)$ into $(ii)$:$-3y = (y-2) + 18$ $\Rightarrow -3y = y + 16$ $\Rightarrow 4y = -16$ $\Rightarrow y = -4$.Substituting $y = -4$ into $(i)$,we get $x = -4-2 = -6$.The intersection point is $(-6, -4)$.The equation of the line with slope $m = \frac{1}{2}$ passing through $(-6, -4)$ is:$y - (-4) = \frac{1}{2}(x - (-6))$ $\Rightarrow y+4 = \frac{1}{2}(x+6)$ $\Rightarrow 2y+8 = x+6$ $\Rightarrow x-2y = 2$.

Let $P(2,4)$ and $Q(18,-12)$ be the points on the parabola $y^2=8x$. The equation of the straight line having slope $\frac{1}{2}$ and passing through the point of intersection of the tangents to the parabola drawn at the points $P$ and $Q$ is

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