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JEE Main 202225 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
25 June
Shift
Evening Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2022 · 25 June · Evening Shift · Mathematics · Application of Derivatives

If the line y = 4 + kx, k > 0 , is the tangent to the parabola y =...

If the line y=4+kx,k>0y = 4 + kx, k > 0, is the tangent to the parabola y=x−x2y = x - {x}^{2} at the point PP and VV is the vertex of the parabola, then the slope of the line through PP and VV isIf the line y = 4 + kx, k > 0 , is the tangent to the parabola y = x - x ^ 2 at the point P and V is the vertex of the

Options

  1. A32\dfrac{3}{2}
  2. B269\dfrac{26}{9}
  3. C52\dfrac{5}{2}
  4. D236\dfrac{23}{6}

Correct answer

C. 52\dfrac{5}{2}

Step-by-step solution

Plotting the given data we have,

If the line y = 4 + kx, k > 0 , is the tangent to the parabola y = x - x ^ 2 at the point P and V is the vertex of the

Now by diagram, Slope of tangent at P=P = Slope of line APAP

(dydx)(α,α−α2)=1−2α=α−α2−4α{\left(\dfrac{dy}{dx}\right)}_{\left(\alpha, \alpha - {\alpha}^{2}\right)} = 1 - 2\alpha = \dfrac{\alpha - {\alpha}^{2} - 4}{\alpha}

On solving we get α=−2⇒P(−2,−6)\alpha = - 2 ⇒ P\left(- 2, - 6\right)

So, slope of PV=52PV = \dfrac{5}{2}

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