Let P : y^{2} = 4ax, a 0 be a parabola with | vedclass

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(D) Let the slope of the line $y = 3x + 5$ be $m_1 = 3$. Let the slopes of the tangents be $m$. The angle between the tangents and the line is $\frac{

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any $a > 0$

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(D) Let the slope of the line $y = 3x + 5$ be $m_1 = 3$. Let the slopes of the tangents be $m$. The angle between the tangents and the line is $\frac{\pi}{4}$.Using the formula $	an(	heta) = |\frac{m - m_1}{1 + m \cdot m_1}|$,we have $	an(\frac{\pi}{4}) = |\frac{m - 3}{1 + 3m}| = 1$.This gives two cases: $\frac{m - 3}{1 + 3m} = 1$ or $\frac{m - 3}{1 + 3m} = -1$.Case $1$: $m - 3 = 1 + 3m \implies -2m = 4 \implies m = -2$.Case $2$: $m - 3 = -1 - 3m \implies 4m = 2 \implies m = \frac{1}{2}$.Since the product of the slopes is $m_1 \cdot m_2 = (-2) \cdot (\frac{1}{2}) = -1$,the two tangents are perpendicular to each other.It is a known property of parabolas that if two tangents are perpendicular,their point of intersection lies on the directrix,and the line segment joining the points of contact $A$ and $B$ passes through the focus $S$.Therefore,$A, S,$ and $B$ are collinear for any $a > 0$.

List-$I$	List-$II$
$A$. Equation of the tangent drawn at $(2, \sqrt{8})$ on the curve $y^2 = 4x$ is	$(i) -36$
$B$. Equation of the normal to the curve $y^2 = 16x$,that makes an angle of $45^{\circ}$ with its axis is	$(ii) 4$
$C$. The chord joining the points $(x_1, y_1)$ and $(x_2, y_2)$ on the curve $y^2 = 12x$ is a focal chord if $y_1 y_2 =$	$(iii) 8$
$D$. $A$ value of $k$ for which $x - 3 = 0$ is the directrix of the curve $y^2 - kx + 16 = 0$ is	$(iv) x - \sqrt{2}y + 2 = 0$
	$(v) x + y - 12 = 0$
	$(vi) x - y - 12 = 0$

Let $P : y^{2} = 4ax, a > 0$ be a parabola with focus $S$. Let the tangents to the parabola $P$ that make an angle of $\frac{\pi}{4}$ with the line $y = 3x + 5$ touch the parabola $P$ at $A$ and $B$. Then the value of $a$ for which $A, B$ and $S$ are collinear is:

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