PQ is a focal chord of the parabola y^2 = 4x | vedclass

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(B) The equation of the parabola is $y^2 = 4x$. Comparing this with $y^2 = 4ax$,we get $a = 1$. The focus $S$ is $(a, 0) = (1, 0)$.Since $P = (4, 4)$

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$\frac{5}{4}$

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(B) The equation of the parabola is $y^2 = 4x$. Comparing this with $y^2 = 4ax$,we get $a = 1$. The focus $S$ is $(a, 0) = (1, 0)$.Since $P = (4, 4)$ lies on the parabola,we can find the parameter $t_1$ for point $P$ using $x = at_1^2$ and $y = 2at_1$. Thus,$4 = 1 \cdot t_1^2 \implies t_1 = 2$.For a focal chord,the parameters $t_1$ and $t_2$ of the endpoints $P$ and $Q$ satisfy $t_1 t_2 = -1$. Therefore,$t_2 = -\frac{1}{t_1} = -\frac{1}{2}$.The focal distance of a point with parameter $t$ on the parabola $y^2 = 4ax$ is given by $a(1 + t^2)$.For point $Q$ with parameter $t_2 = -\frac{1}{2}$,the focal distance $SQ = a(1 + t_2^2) = 1 \cdot (1 + (-\frac{1}{2})^2) = 1 + \frac{1}{4} = \frac{5}{4}$.

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	$(iv) x - \sqrt{2}y + 2 = 0$
	$(v) x + y - 12 = 0$
	$(vi) x - y - 12 = 0$

$PQ$ is a focal chord of the parabola $y^2 = 4x$ with focus $S$. If $P = (4, 4)$,then $SQ = $

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