If y^2=16x is the given parabola,then the poi | vedclass

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(B) Given parabola is $y^2=16x$. Comparing with $y^2=4ax$,we get $a=4$. Thus,the focus is $(4,0)$.The focal chord passes through $(2,2)$ and $(4,0)$.

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$(9,-5)$

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(B) Given parabola is $y^2=16x$. Comparing with $y^2=4ax$,we get $a=4$. Thus,the focus is $(4,0)$.The focal chord passes through $(2,2)$ and $(4,0)$. The slope $m = \frac{0-2}{4-2} = -1$.The equation of the focal chord is $y-0 = -1(x-4)$,which simplifies to $y = -x+4$ or $x+y=4$.The double ordinate has length $24$. For a parabola $y^2=16x$,the double ordinate is a line perpendicular to the axis of symmetry. If the length is $24$,then $2|y| = 24$,so $y = 12$ or $y = -12$.Substituting $y=12$ into $y^2=16x$,we get $144=16x$,so $x=9$. The point is $(9,12)$.Substituting $y=-12$ into $y^2=16x$,we get $144=16x$,so $x=9$. The point is $(9,-12)$.We need the intersection of the focal chord $x+y=4$ and the double ordinate $x=9$.Substituting $x=9$ into $x+y=4$,we get $9+y=4$,which gives $y=-5$.Thus,the point of intersection is $(9,-5)$.

If $y^2=16x$ is the given parabola,then the point of intersection of the focal chord passing through the point $(2,2)$ and the double ordinate of length $24$ is

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