Let O be the vertex of the parabola x^{2}=4y | vedclass

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(D) Let $Q = (2t, t^2)$ be a point on the parabola $x^2 = 4y$. The vertex $O$ is $(0, 0)$.The point $P(h, k)$ divides $OQ$ in the ratio $2:3$. Using t

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$5x - 4y + 3 = 0$

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(D) Let $Q = (2t, t^2)$ be a point on the parabola $x^2 = 4y$. The vertex $O$ is $(0, 0)$.The point $P(h, k)$ divides $OQ$ in the ratio $2:3$. Using the section formula:$h = \frac{2(2t) + 3(0)}{2+3} = \frac{4t}{5} \Rightarrow t = \frac{5h}{4}$$k = \frac{2(t^2) + 3(0)}{2+3} = \frac{2t^2}{5} = \frac{2}{5} \left(\frac{5h}{4}\right)^2 = \frac{2}{5} \cdot \frac{25h^2}{16} = \frac{5h^2}{8}$Thus, the locus $C$ is $8k = 5h^2$, or $5x^2 = 8y$.The equation of the chord of the parabola $5x^2 = 8y$ bisected at $(x_1, y_1) = (1, 2)$ is given by $T = S_1$, where $T = 5x(x_1) - 4(y + y_1)$ and $S_1 = 5x_1^2 - 8y_1$.Substituting the values:$5x(1) - 4(y + 2) = 5(1)^2 - 8(2)$$5x - 4y - 8 = 5 - 16$$5x - 4y - 8 = -11$$5x - 4y + 3 = 0$

Let $O$ be the vertex of the parabola $x^{2}=4y$ and $Q$ be any point on it. Let the locus of the point $P$, which divides the line segment $OQ$ internally in the ratio $2:3$, be the conic $C$. Then the equation of the chord of $C$, which is bisected at the point $(1, 2)$, is:

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