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Q.
The vertex of the parabola $2\left((x-1)^{2}+(y-2)^{2}\right)=(x+y$ $+3)^{2}$ is
Conic Sections
Solution:
$2\left((x-1)^{2}+(y-2)^{2}\right)=(x+y+3)^{2}$
$\Rightarrow \sqrt{(x-1)^{2}+(y-2)^{2}}=\frac{|x+y+3|}{\sqrt{2}}$
So, focus is $S(1,2)$ and directrix is $x+y+3=0$.
Axis of the parabola is $x-y+1=0$.
Solving directrix and axis, we get foot of perpendicular of directrix on axis as $A(-2,-1)$.
Therefore, vertex is mid-point of $A S$ which is $\left(-\frac{1}{2}, \frac{1}{2}\right)$.