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

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(C) The given equation of the parabola is $y^2 = kx - 8$,which can be rewritten as $y^2 = k(x - 8/k)$.Comparing this with the standard form $Y^2 = 4AX

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$4$

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(C) The given equation of the parabola is $y^2 = kx - 8$,which can be rewritten as $y^2 = k(x - 8/k)$.Comparing this with the standard form $Y^2 = 4AX$,where $Y = y$ and $X = x - 8/k$,we have $4A = k$,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 $32 - k^2 = 4k$,which leads to the quadratic equation $k^2 + 4k - 32 = 0$.Factoring the quadratic,we get $(k + 8)(k - 4) = 0$.Thus,the possible values for $k$ are $k = -8$ or $k = 4$.

If the line $x - 1 = 0$ is the directrix of the parabola $y^2 - kx + 8 = 0$,then what is one value of $k$?

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