The focal distance of a point on the parabola | vedclass

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Question

(B) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $4a = 16$,so $a = 4$.Let the point on the parabola be $(h, k)$

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) The equation of the parabola is $y^2 = 16x$. Comparing this 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 $k = 2h$.Since the point lies on the parabola,$k^2 = 16h$.Substituting $k = 2h$ into the equation,we get $(2h)^2 = 16h$,which simplifies to $4h^2 = 16h$.This gives $4h^2 - 16h = 0$,or $4h(h - 4) = 0$.Thus,$h = 0$ or $h = 4$.If $h = 0$,then $k = 0$. The point is $(0, 0)$. The focal distance is $h + a = 0 + 4 = 4$.If $h = 4$,then $k = 8$. The point is $(4, 8)$. The focal distance is $h + a = 4 + 4 = 8$.Since $4$ is not among the options,the required focal distance 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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