Let the locus of the mid-point of the chord t | vedclass

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(B) Let the parabola be $y^2=4x$. Any point on the parabola is $P(t^2, 2t)$.The chord passes through the origin $O(0,0)$ and $P(t^2, 2t)$.The mid-poin

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$2y^2=3x$

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(B) Let the parabola be $y^2=4x$. Any point on the parabola is $P(t^2, 2t)$.The chord passes through the origin $O(0,0)$ and $P(t^2, 2t)$.The mid-point $M(h, k)$ of the chord $OP$ is given by $h = \frac{t^2+0}{2} = \frac{t^2}{2}$ and $k = \frac{2t+0}{2} = t$.Substituting $t=k$ into $h = \frac{t^2}{2}$, we get $h = \frac{k^2}{2}$, so $k^2=2h$.The locus $S$ is $y^2=2x$.Let $P(x_0, y_0)$ be a point on $S$, so $y_0^2=2x_0$. Since $P$ is on $S$, we can write $P$ as $(2t^2, 2t)$ because $(2t)^2 = 2(2t^2)$.Let $R(x, y)$ be the point that divides $OP$ internally in the ratio $3:1$. Using the section formula:$x = \frac{3(2t^2) + 1(0)}{3+1} = \frac{6t^2}{4} = \frac{3t^2}{2}$$y = \frac{3(2t) + 1(0)}{3+1} = \frac{6t}{4} = \frac{3t}{2}$From $y = \frac{3t}{2}$, we get $t = \frac{2y}{3}$.Substituting $t$ into $x = \frac{3t^2}{2}$:$x = \frac{3}{2} \left(\frac{2y}{3}\right)^2 = \frac{3}{2} \cdot \frac{4y^2}{9} = \frac{2y^2}{3}$Thus, $3x = 2y^2$, or $2y^2=3x$.

Let the locus of the mid-point of the chord through the origin $O$ of the parabola $y^{2}=4x$ be the curve $S$. Let $P$ be any point on $S$. Then the locus of the point, which internally divides $OP$ in the ratio $3:1$, is:

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