If the line y = 4 + kx,k 0,is the tangent t | vedclass

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(C) The equation of the parabola is $y = x - x^{2}$.Let the point of tangency be $P(\alpha, \alpha - \alpha^{2})$.The slope of the tangent at $P$ is g

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$\frac{5}{2}$

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(C) The equation of the parabola is $y = x - x^{2}$.Let the point of tangency be $P(\alpha, \alpha - \alpha^{2})$.The slope of the tangent at $P$ is given by $\frac{dy}{dx} = 1 - 2x$. At $x = \alpha$,the slope is $1 - 2\alpha$.The line $y = kx + 4$ passes through $A(0, 4)$ and $P(\alpha, \alpha - \alpha^{2})$.The slope of the line $AP$ is $\frac{(\alpha - \alpha^{2}) - 4}{\alpha - 0} = \frac{\alpha - \alpha^{2} - 4}{\alpha}$.Equating the slopes: $1 - 2\alpha = \frac{\alpha - \alpha^{2} - 4}{\alpha}$.$\alpha(1 - 2\alpha) = \alpha - \alpha^{2} - 4$$\alpha - 2\alpha^{2} = \alpha - \alpha^{2} - 4$$\alpha^{2} = 4 \Rightarrow \alpha = \pm 2$.Since $k > 0$,the slope $1 - 2\alpha$ must be positive,so $1 - 2\alpha > 0 \Rightarrow \alpha The point $P$ is $(-2, -2 - (-2)^{2}) = (-2, -6)$.The vertex $V$ of the parabola $y = -(x^{2} - x) = -(x - \frac{1}{2})^{2} + \frac{1}{4}$ is $(\frac{1}{2}, \frac{1}{4})$.The slope of the line through $P(-2, -6)$ and $V(\frac{1}{2}, \frac{1}{4})$ is $\frac{\frac{1}{4} - (-6)}{\frac{1}{2} - (-2)} = \frac{\frac{25}{4}}{\frac{5}{2}} = \frac{25}{4} 	imes \frac{2}{5} = \frac{5}{2}$.

List-$I$ (Geometric Property)	List-$II$ (Coordinates/Equations)
$I$. Vertex	$A$. $\left(-\frac{3}{2}, -3\right)$
$II$. Focus	$B$. $\left(\frac{3}{2}, -3\right)$
$III$. Equation of the directrix	$C$. $2x + 5 = 0$
$IV$. Equation of the axis	$D$. $2x + y + 3 = 0$
	$E$. $y + 3 = 0$
	$F$. $(-2, -3)$

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

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