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

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(A) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $4a = 16$,so $a = 4$.The focus of the parabola is $S(a, 0) = (

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$25$

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(A) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $4a = 16$,so $a = 4$.The focus of the parabola is $S(a, 0) = (4, 0)$.Let the coordinates of one end of the focal chord be $A(at_1^2, 2at_1) = (1, 4)$.Since $a = 4$,we have $4t_1^2 = 1$ $\Rightarrow t_1^2 = \frac{1}{4}$ $\Rightarrow t_1 = \frac{1}{2}$ (since $y > 0$).For a focal chord,the product of the parameters of the endpoints is $t_1 t_2 = -1$. Thus,$t_2 = -\frac{1}{t_1} = -2$.The length of a focal chord with parameter $t$ is given by $L = a(t + \frac{1}{t})^2$.Substituting $t = t_1 = \frac{1}{2}$,we get:$L = 4(\frac{1}{2} + \frac{1}{1/2})^2 = 4(\frac{1}{2} + 2)^2 = 4(\frac{5}{2})^2 = 4 	imes \frac{25}{4} = 25$.

If one end of a focal chord of the parabola $y^2 = 16x$ is at $(1, 4)$,then the length of this focal chord is

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