JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Parabola

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 :

Answer: (C)

Solution

The parabola is

y = x - x^{2}

. Its derivative is

y^{'} = 1 - 2 x

. At a point

(α, α - α^{2})

, the slope of the tangent is

1 - 2 α

. The tangent line is

y - (α - α^{2}) = (1 - 2 α) (x - α)

, which simplifies to

y = (1 - 2 α) x + α^{2}

. Comparing with

y = k x + 4

, we have

k = 1 - 2 α

and

α^{2} = 4

, so

α = \pm 2

. Since

k > 0

, we need

1 - 2 α > 0

, so

α < \frac{1}{2}

. Thus

α = - 2

, giving

P = (- 2, - 6)

. The vertex of

y = x - x^{2} = - (x^{2} - x) = - (x - \frac{1}{2})^{2} + \frac{1}{4}

V = (\frac{1}{2}, \frac{1}{4})

. The slope of PV is

\frac{\frac{1}{4} - ( - 6 )}{\frac{1}{2} - ( - 2 )} = \frac{\frac{25}{4}}{\frac{5}{2}} = \frac{25}{4} \cdot \frac{2}{5} = \frac{5}{2}

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option