If P is a point on the parabola y^2=8x and A | vedclass

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(A) Let $P$ be a point on the parabola $y^2=8x$. The coordinates of $P$ can be represented as $(2t^2, 4t)$.Let $M(x, y)$ be the mid-point of the line

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$y^2=4(x-\frac{1}{2})$

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(A) Let $P$ be a point on the parabola $y^2=8x$. The coordinates of $P$ can be represented as $(2t^2, 4t)$.Let $M(x, y)$ be the mid-point of the line segment $AP$,where $A$ is $(1, 0)$.Using the mid-point formula,we have:$x = \frac{2t^2 + 1}{2}$ and $y = \frac{4t + 0}{2} = 2t$.From $y = 2t$,we get $t = \frac{y}{2}$.Substituting $t$ into the equation for $x$:$x = \frac{2(\frac{y}{2})^2 + 1}{2} = \frac{2(\frac{y^2}{4}) + 1}{2} = \frac{\frac{y^2}{2} + 1}{2} = \frac{y^2 + 2}{4}$.$4x = y^2 + 2$.$y^2 = 4x - 2 = 4(x - \frac{1}{2})$.Thus,the locus of the mid-point is $y^2 = 4(x - \frac{1}{2})$.

If $P$ is a point on the parabola $y^2=8x$ and $A$ is the point $(1,0)$,then the locus of the mid-point of the line segment $AP$ is

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