Question

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 1) \( \frac{3}{2} \) 2) \( \frac{26}{9} \) 3) \( \frac{5}{2} \) 4) \( \frac{23}{6} \)

Answer 1

Slope of tangent at P = Slope of line AP

\(y'|_P=1-2\alpha=\frac{\alpha-\alpha^2-4}{\alpha-0}\)

Solving α = -2 ⇒ P(-2, -6), 

vertex of the parabola = \((\frac{1}{2},\frac{1}{4})\)

Slope of PV = \(\frac{5}{2}\)