The focal distance of a point on the parabola | vedclass

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(B) The given parabola is $y^{2}=16x$. Comparing with $y^{2}=4ax$,we get $4a=16$,so $a=4$. Let the point on the parabola be $(h, k)$. Given that the o

Vedclass · Accepted Answer

$8$

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(B) The given parabola is $y^{2}=16x$. Comparing with $y^{2}=4ax$,we get $4a=16$,so $a=4$. Let the point on the parabola be $(h, k)$. Given that the ordinate is twice the abscissa,we have $k=2h$. Substituting $k=2h$ into the parabola equation: $(2h)^{2}=16h$ $\Rightarrow 4h^{2}=16h$ $\Rightarrow 4h(h-4)=0$. Thus,$h=0$ or $h=4$. For $h=0$,$k=0$. For $h=4$,$k=8$. The point $(0,0)$ is the vertex,where the focal distance is $a=4$. The point $(4,8)$ is on the parabola,where the focal distance is $h+a = 4+4 = 8$. Since the question asks for the focal distance of a point (implying a non-vertex point),the answer is $8$.

The focal distance of a point on the parabola $y^{2}=16x$ whose ordinate is twice the abscissa,is

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