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If the line $y = 4 + k x, k > 0,$ is the tangent to the parabola $y = x - x^{2}$ at the point P and V is the vertex of the parabola, then the slope of the line through P and V is :

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$\frac{26}{9}$

$\frac{23}{6}$

$\frac{5}{2}$

$\frac{3}{2}$

answer is C.

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Detailed Solution

Slope of tangent at P= Slope of line AP

${y^{'}|}_{P} = 1 - 2 α = \frac{α - α^{2} - 4}{α - 0}$

Solving $α = - 2 \Rightarrow P (- 2, - 6)$ , vertex of the parabola= $(\frac{1}{2}, \frac{1}{4})$

Slope of PV = $\frac{5}{2}$

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