If the focal chord drawn through the point (1 | vedclass

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(C) The given parabola is $y^2=8x$,which is of the form $y^2=4ax$,so $a=2$. The focus is $(a, 0) = (2, 0)$.Let the points on the parabola be $(x_1, y_

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$6$

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(C) The given parabola is $y^2=8x$,which is of the form $y^2=4ax$,so $a=2$. The focus is $(a, 0) = (2, 0)$.Let the points on the parabola be $(x_1, y_1) = (at_1^2, 2at_1)$ and $(x_2, y_2) = (at_2^2, 2at_2)$.Since the chord is a focal chord,it passes through the focus $(2, 0)$,which implies $t_1 t_2 = -1$.The chord also passes through the point $(1, 2)$. The equation of the chord joining $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$ is $y(t_1+t_2) = 2x + 2at_1 t_2$.Substituting $a=2$ and $t_1 t_2 = -1$,we get $y(t_1+t_2) = 2x - 4$.Since the chord passes through $(1, 2)$,we have $2(t_1+t_2) = 2(1) - 4 = -2$,so $t_1+t_2 = -1$.We need to find $x_1+x_2 = at_1^2 + at_2^2 = a(t_1^2 + t_2^2) = a((t_1+t_2)^2 - 2t_1 t_2)$.Substituting the values $a=2$,$t_1+t_2 = -1$,and $t_1 t_2 = -1$:$x_1+x_2 = 2((-1)^2 - 2(-1)) = 2(1+2) = 2(3) = 6$.

If the focal chord drawn through the point $(1,2)$ to the parabola $y^2=8x$ meets this parabola in $(x_1, y_1)$ and $(x_2, y_2)$,then $x_1+x_2=$

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