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# The point of intersection of the tangents at $t_{1}=t$ and $t_{2}=3t$ to the parabola $y^{2}=8x$ is

$\begin{array}{1 1}(1)(6t^{2}, 8t)&(2)(8t,6t^{2})\\(3)(t^{2},4t)&(4)(4t,t^{2})\end{array}$

The given equation is $y^2=8x$
$y^2=4(2)x =>a=2$
Given $t_1=t,t_2=3t$
$\therefore$ The point of intersection is
$\{2 \times t \times 3t , 2(t +3t)\}$
$(6t^{2}, 8t)$
Hence 1 is the correct answer