If x = 1 is the directrix of the parabola y<sup>2</sup> = kx - 8, then k is:
Step-by-step Solution:
1. Given Equation of Parabola: \[ y^2 - kx + 8 = 0 \] Rearrange the equation: \[ y^2 = kx - 8 \] \[ y^2 = k\left(x - \frac{8}{k}\right) \] This is of the form: \[ (y - 0)^2 = k\left(x - \frac{8}{k}\right) \] 2. Identify the Directrix: For a parabola of the form \( y^2 = 4a(x - h) \), the directrix is \( x = h - a \). Comparing with the given equation: \[ 4a = k \Rightarrow a = \frac{k}{4} \] The directrix is: \[ x = \frac{8}{k} - \frac{k}{4} \] 3. Given Directrix: The directrix is given as \( x - 1 = 0 \) or \( x = 1 \). Therefore: \[ \frac{8}{k} - \frac{k}{4} = 1 \] 4. Solve for \( k \): Multiply through by \( 4k \) to eliminate the denominators: \[ 32 - k^2 = 4k \] Rearrange the equation: \[ k^2 + 4k - 32 = 0 \] Solve the quadratic equation: \[ k = \frac{-4 \pm \sqrt{16 + 128}}{2} = \frac{-4 \pm \sqrt{144}}{2} = \frac{-4 \pm 12}{2} \] \[ k = \frac{8}{2} = 4 \quad \text{or} \quad k = \frac{-16}{2} = -8 \] Therefore, the possible values of \( k \) are: \[ \boxed{4 \quad \text{or} \quad -8} \]