TP and TQ are tangents of the parabola y^2 = | vedclass

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(C) Let the coordinates of point $T$ be $(h, k)$.The equation of the chord of contact $PQ$ for the parabola $y^2 = 4ax$ from point $(h, k)$ is given b

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$- \frac{4}{3}$

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(C) Let the coordinates of point $T$ be $(h, k)$.The equation of the chord of contact $PQ$ for the parabola $y^2 = 4ax$ from point $(h, k)$ is given by $ky = 2a(x + h)$.For the given parabola $y^2 = 8x$,we have $4a = 8$,so $a = 2$.Thus,the equation of the chord of contact is $ky = 2(2)(x + h)$,which simplifies to $ky = 4(x + h)$.Since the chord $PQ$ passes through the point $(-2, 3)$,we substitute $x = -2$ and $y = 3$ into the equation:$3k = 4(-2 + h)$$3k = 4h - 8$Replacing $(h, k)$ with $(x, y)$ to find the locus of $T$:$3y = 4x - 8$$y = \frac{4}{3}x - \frac{8}{3}$Comparing this with $y = mx + c$,we get $m = \frac{4}{3}$ and $c = -\frac{8}{3}$.Therefore,$m + c = \frac{4}{3} - \frac{8}{3} = -\frac{4}{3}$.

$TP$ and $TQ$ are tangents of the parabola $y^2 = 8x$ at $P$ and $Q$ respectively. If the chord $PQ$ passes through the point $(-2, 3)$ and the locus of point $T$ is $y = mx + c$,then $(m + c)$ is equal to -

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