If one end of a focal chord of the parabola y | vedclass

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(D) The equation of the parabola is $y^2 = \frac{8}{a} x$. Since the point $(1, 4)$ lies on the parabola,we substitute $x = 1$ and $y = 4$: $4^2 = \fr

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$25$

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(D) The equation of the parabola is $y^2 = \frac{8}{a} x$. Since the point $(1, 4)$ lies on the parabola,we substitute $x = 1$ and $y = 4$: $4^2 = \frac{8}{a} (1)$ $\Rightarrow 16 = \frac{8}{a}$ $\Rightarrow a = \frac{1}{2}$. Substituting $a = \frac{1}{2}$ into the equation,we get $y^2 = \frac{8}{1/2} x = 16x$. Comparing $y^2 = 16x$ with $y^2 = 4Ax$,we find $4A = 16$,so $A = 4$. The focus of the parabola $y^2 = 4Ax$ is $(A, 0)$,which is $(4, 0)$. Let the endpoints of the focal chord be $P(x_1, y_1)$ and $Q(x_2, y_2)$. Given $P = (1, 4)$. For a parabola $y^2 = 4Ax$,if one end is $(At_1^2, 2At_1)$,the other end is $(At_2^2, 2At_2)$ where $t_1 t_2 = -1$. Here $A = 4$,so $4t_1^2 = 1$ $\Rightarrow t_1^2 = \frac{1}{4}$ $\Rightarrow t_1 = -\frac{1}{2}$ (since $y_1 = 2At_1 = 8t_1 = 4$). Then $t_2 = -\frac{1}{t_1} = 2$. The coordinates of $Q$ are $(A t_2^2, 2A t_2) = (4(2^2), 2(4)(2)) = (16, 16)$. The length of the focal chord is the distance between $(1, 4)$ and $(16, 16)$: $L = \sqrt{(16 - 1)^2 + (16 - 4)^2} = \sqrt{15^2 + 12^2} = \sqrt{225 + 144} = \sqrt{369} = 3\sqrt{41}$. Wait,re-evaluating the focal chord property: The length of a focal chord with parameter $t$ is $A(t + 1/t)^2$. Here $t = -1/2$,so $L = 4(-1/2 - 2)^2 = 4(-5/2)^2 = 4(25/4) = 25$.

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 one end of a focal chord of the parabola $y^2 = \frac{8}{a} x$ $(a > 0)$ is at $(1, 4)$,then the length of this focal chord is

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