If the locus of a point which divides a chord | vedclass

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(D) Let $P(t_1^2, 2t_1)$ and $Q(t_2^2, 2t_2)$ be the endpoints of the chord.Let the point $R(h, k)$ divide the chord $PQ$ internally in the ratio $1:3

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$\left(\frac{3}{16}, \frac{3}{4}\right)$

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(D) Let $P(t_1^2, 2t_1)$ and $Q(t_2^2, 2t_2)$ be the endpoints of the chord.Let the point $R(h, k)$ divide the chord $PQ$ internally in the ratio $1:3$.Using the section formula,we have:$h = \frac{1 \cdot t_2^2 + 3 \cdot t_1^2}{1+3} = \frac{t_2^2 + 3t_1^2}{4}$$k = \frac{1 \cdot 2t_2 + 3 \cdot 2t_1}{1+3} = \frac{2t_2 + 6t_1}{4} = \frac{t_2 + 3t_1}{2}$The slope of the chord $PQ$ is given as $2$:$\frac{2t_2 - 2t_1}{t_2^2 - t_1^2} = 2$$\frac{2(t_2 - t_1)}{(t_2 - t_1)(t_2 + t_1)} = 2$$\frac{2}{t_2 + t_1} = 2 \Rightarrow t_1 + t_2 = 1 \Rightarrow t_2 = 1 - t_1$Substitute $t_2 = 1 - t_1$ into the expressions for $h$ and $k$:$k = \frac{(1 - t_1) + 3t_1}{2} = \frac{2t_1 + 1}{2} = t_1 + \frac{1}{2} \Rightarrow t_1 = k - \frac{1}{2}$$4h = 3t_1^2 + (1 - t_1)^2 = 3t_1^2 + 1 - 2t_1 + t_1^2 = 4t_1^2 - 2t_1 + 1$Substitute $t_1 = k - \frac{1}{2}$ into the equation for $4h$:$4h = 4(k - \frac{1}{2})^2 - 2(k - \frac{1}{2}) + 1$$4h = 4(k^2 - k + \frac{1}{4}) - 2k + 1 + 1$$4h = 4k^2 - 4k + 1 - 2k + 2 = 4k^2 - 6k + 3$$h = k^2 - \frac{3}{2}k + \frac{3}{4}$$k^2 - \frac{3}{2}k = h - \frac{3}{4}$$(k - \frac{3}{4})^2 = h - \frac{3}{4} + \frac{9}{16} = h - \frac{3}{16}$Comparing with $(y - k_0)^2 = 4a(x - h_0)$,the vertex is $\left(\frac{3}{16}, \frac{3}{4}\right)$.

If the locus of a point which divides a chord with slope $2$ of the parabola $y^2=4x$ internally in the ratio $1:3$ is a parabola,then its vertex is

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