The line x - 1 = 0 is the directrix of the pa | vedclass

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(C) The given equation of the parabola is ${y^2} - kx + 8 = 0$,which can be rewritten as ${y^2} = k(x - 8/k)$.This is in the form ${Y^2} = 4AX$,where

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$4$

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(C) The given equation of the parabola is ${y^2} - kx + 8 = 0$,which can be rewritten as ${y^2} = k(x - 8/k)$.This is in the form ${Y^2} = 4AX$,where $Y = y$,$X = x - 8/k$,and $4A = k$,so $A = k/4$.The directrix of the parabola ${Y^2} = 4AX$ is $X + A = 0$.Substituting the values of $X$ and $A$,we get $(x - 8/k) + k/4 = 0$,which simplifies to $x = 8/k - k/4$.We are given that the directrix is $x - 1 = 0$,or $x = 1$.Equating the two expressions for $x$,we have $8/k - k/4 = 1$.Multiplying by $4k$,we get $32 - k^2 = 4k$,or ${k^2} + 4k - 32 = 0$.Factoring the quadratic equation,we get $(k + 8)(k - 4) = 0$.Thus,the values of $k$ are $k = -8$ or $k = 4$.

The line $x - 1 = 0$ is the directrix of the parabola ${y^2} - kx + 8 = 0$. Then one of the values of $k$ is

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