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Let $\mathrm{Z}$ be the point of intersection of the axis and the directrix of the parabola $4 x^2-12 x+4 y+5=0$. If $S$ is its focus, then the point which divides SZ in the ratio $2: 1$ is
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$\left(\frac{3}{2}, \frac{13}{12}\right)$
Given equation of parabola is $4 x^2-12 x+4 y+5=0$
$\begin{aligned} & \text { i.e }\left(x-\frac{3}{2}\right)^2=-(y-1) \Rightarrow \text { vertex }=\left(\frac{3}{2}, 1\right) \\ & \therefore Z=\left(0+\frac{3}{2}, \frac{1}{4}+1\right)=\left(\frac{3}{2}, \frac{5}{4}\right) \\ & \text { and focus }(s)=\left(0+\frac{3}{2},-\frac{1}{4}+1\right)=\left(\frac{3}{2}, \frac{3}{4}\right) \\ & \text { Now, required point }=\left(\frac{6}{\frac{2}{2}+\frac{3}{2}}, \frac{10}{4}+\frac{3}{4}\right)=\left(\frac{3}{2}, \frac{13}{12}\right)\end{aligned}$
$\begin{aligned} & \text { i.e }\left(x-\frac{3}{2}\right)^2=-(y-1) \Rightarrow \text { vertex }=\left(\frac{3}{2}, 1\right) \\ & \therefore Z=\left(0+\frac{3}{2}, \frac{1}{4}+1\right)=\left(\frac{3}{2}, \frac{5}{4}\right) \\ & \text { and focus }(s)=\left(0+\frac{3}{2},-\frac{1}{4}+1\right)=\left(\frac{3}{2}, \frac{3}{4}\right) \\ & \text { Now, required point }=\left(\frac{6}{\frac{2}{2}+\frac{3}{2}}, \frac{10}{4}+\frac{3}{4}\right)=\left(\frac{3}{2}, \frac{13}{12}\right)\end{aligned}$
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