JEE Main 2022 · 25 Jun · Shift 2 · Mathematics Q18

← JEE Main 2022 · 25 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 18 of 90

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 parabola, then the slope of the line through $P$ and $V$ is:

\dfrac32

\dfrac{26}9

\dfrac52

\dfrac{23}6

Answer and worked solution

Official answer: C

Tangency. $x-x^2=4+kx$ gives $x^2+(k-1)x+4=0$ with zero discriminant: $(k-1)^2=16$ , $k=5$ .

Point. $x=-2$ , $y=-6$ : $P(-2,-6)$ . Vertex $V\left(\frac12,\frac14\right)$ .

Slope. $\dfrac{\frac14+6}{\frac12+2}=\dfrac{25/4}{5/2}=\dfrac52$ .

(C)

Topics:Parabola Tangents and normals Nature of roots

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JEE Main 2022 · 25 Jun · Shift 2, Mathematics question 18