If one end of a focal chord AB of the parabol | vedclass

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(B) The equation of the parabola is $y^{2}=8x$,which is of the form $y^{2}=4ax$,where $4a=8$,so $a=2$.The coordinates of any point on the parabola are

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$x-2y+8=0$

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(B) The equation of the parabola is $y^{2}=8x$,which is of the form $y^{2}=4ax$,where $4a=8$,so $a=2$.The coordinates of any point on the parabola are $(at^{2}, 2at) = (2t^{2}, 4t)$.For point $A\left(\frac{1}{2}, -2\right)$,we have $4t_{1}=-2$,which gives $t_{1}=-\frac{1}{2}$.For a focal chord,the product of the parameters of the endpoints is $t_{1}t_{2}=-1$.Substituting $t_{1}=-\frac{1}{2}$,we get $t_{2} = -\frac{1}{t_{1}} = -\frac{1}{-1/2} = 2$.The coordinates of point $B$ are $(2t_{2}^{2}, 4t_{2}) = (2(2)^{2}, 4(2)) = (8, 8)$.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 $B(8, 8)$ and $a=2$,the equation of the tangent is $y(8) = 2(2)(x+8)$.$8y = 4(x+8)$ $\Rightarrow 2y = x+8$ $\Rightarrow x-2y+8=0$.

If one end of a focal chord $AB$ of the parabola $y^{2}=8x$ is at $A\left(\frac{1}{2},-2\right)$,then the equation of the tangent to it at $B$ is:

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