A point P(x, y) moves such that its distance | vedclass

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(A) Let the point be $P(x, y)$.The distance of $P(x, y)$ from the point $(a, 0)$ is $\sqrt{(x - a)^2 + (y - 0)^2}$.The distance of $P(x, y)$ from the

Vedclass · Accepted Answer

$y^2 = 4ax$

Vedclass · Answer

(A) Let the point be $P(x, y)$.The distance of $P(x, y)$ from the point $(a, 0)$ is $\sqrt{(x - a)^2 + (y - 0)^2}$.The distance of $P(x, y)$ from the line $x + a = 0$ is $|x + a|$.According to the problem,these distances are equal:$\sqrt{(x - a)^2 + y^2} = |x + a|$Squaring both sides,we get:$(x - a)^2 + y^2 = (x + a)^2$$x^2 - 2ax + a^2 + y^2 = x^2 + 2ax + a^2$$y^2 = 4ax$This is the standard equation of a parabola.

$A$ point $P(x, y)$ moves such that its distance from the point $(a, 0)$ is always equal to its distance from the line $x + a = 0$. The locus of the point is:

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