JEE Main 2022 (Online) 25th June Evening ShiftQuestion 1
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:
Parabola
Solution
Concept. Tangency means a double root; the double root gives the point of contact. Formulas used. (1) $x^2+Bx+C=0$ with $B^2=4C$ has the double root $x=-\dfrac B2$.(2) Vertex of $y=-(x-h)^2+k$ is $(h,k)$.Step 1 - Substitute. $x-x^2=4+kx\Rightarrow x^2+(k-1)x+4=0$. Step 2 - Use (1). $(k-1)^2=16\Rightarrow k=5$ or $-3$; $k>0$ gives $k=5$. Contact: $x=-\frac{4}{2}=-2$, $y=-2-4=-6$. $P(-2,-6)$. Step 3 - Use (2). $y=-\big(x-\frac12\big)^2+\frac14$: $V\big(\frac12,\frac14\big)$. Step 4 - Slope. $\dfrac{\frac14+6}{\frac12+2}=\dfrac{25/4}{5/2}=\dfrac52$. Answer: C