Let O be the vertex and Q be any point on the | vedclass

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(D) The vertex of the parabola $x^2=8y$ is $O(0, 0)$.Let the coordinates of point $Q$ be $(x_1, y_1)$. Since $Q$ lies on the parabola, $x_1^2 = 8y_1$.

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$x^2=2y$

Vedclass · Answer

(D) The vertex of the parabola $x^2=8y$ is $O(0, 0)$.Let the coordinates of point $Q$ be $(x_1, y_1)$. Since $Q$ lies on the parabola, $x_1^2 = 8y_1$.Let the coordinates of point $P$ be $(h, k)$.Since $P$ divides $OQ$ in the ratio $1:3$, by the section formula:$h = \frac{1 \cdot x_1 + 3 \cdot 0}{1+3} = \frac{x_1}{4} \Rightarrow x_1 = 4h$$k = \frac{1 \cdot y_1 + 3 \cdot 0}{1+3} = \frac{y_1}{4} \Rightarrow y_1 = 4k$Substituting these into the parabola equation $x_1^2 = 8y_1$:$(4h)^2 = 8(4k)$$16h^2 = 32k$$h^2 = 2k$Replacing $(h, k)$ with $(x, y)$, the locus of $P$ is $x^2 = 2y$.

Let $O$ be the vertex and $Q$ be any point on the parabola $x^2=8y$. If the point $P$ divides the line segment $OQ$ internally in the ratio $1:3$, then the locus of $P$ is

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