Let PQ be a chord of the parabola y^2=12x and | vedclass

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(D) The equation of a chord of a parabola $y^2=4ax$ with midpoint $(x_1, y_1)$ is given by $T=S_1$.Here,the parabola is $y^2=12x$,so $4a=12$,which mea

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$\left(\frac{1}{2},-20\right)$

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(D) The equation of a chord of a parabola $y^2=4ax$ with midpoint $(x_1, y_1)$ is given by $T=S_1$.Here,the parabola is $y^2=12x$,so $4a=12$,which means $a=3$.The midpoint $(x_1, y_1)$ is $(4, 1)$.The equation of the chord is $yy_1 - 2a(x+x_1) = y_1^2 - 4ax_1$.Substituting the values,we get $y(1) - 2(3)(x+4) = (1)^2 - 12(4)$.$y - 6(x+4) = 1 - 48$.$y - 6x - 24 = -47$.$6x - y = 23$.Now,we check which of the given points satisfies the equation $6x - y = 23$.For option $D$: $6\left(\frac{1}{2}\right) - (-20) = 3 + 20 = 23$.Thus,the point $\left(\frac{1}{2}, -20\right)$ lies on the line.

Let $PQ$ be a chord of the parabola $y^2=12x$ and the midpoint of $PQ$ be at $(4,1)$. Then,which of the following points lies on the line passing through the points $P$ and $Q$?

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