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Q.
The equation of the parabola whose focus $(3,2)$ and vertex $(1,2)$, is
Conic Sections
Solution:
Let $(x_{1}, y_{1})$ be the coordinates of the point of intersection of the axis and the directrix. Then the vertex is the mid point of the line segment joining $(x_{1}, y_{1})$ and focus $(3,2)$
$\Rightarrow \, \frac{x_{1}+3}{2}=1$
$\Rightarrow x_{1}+3=2$
$\Rightarrow x_{1}=-1$
and $\frac{y_{1}+2}{2}=2$
$\Rightarrow y_{1}+2=4$
$\Rightarrow y_{1}=2$
$\Rightarrow $ The directrix meets the axis at $\left(- 1, 2\right)$
Let $m_{1}$ be the slope of the axis.
Then $m_{1} $= slope of the line joining the focus and the vertex $=\frac{2-2}{3-1}=\frac{0}{2}$
$\therefore $ Directrix passes through $\left(-1 ,2 \right)$ and have
slope $\frac{-2}{0} \therefore x=-1$ is the directrix.
$\therefore $ By definition, equation of parabola is
$\sqrt{\left(x-3\right)^{2}+\left(y-2\right)^{2}}=\left(x+1\right)$
$\Rightarrow \left(x-3\right)^{2}+\left(y-2\right)^{2}=\left(x+1\right)^{2}$
$\Rightarrow y^{2} - 4y -8x+ 12 = 0$