Question

The vertex of the parabola $2\left((x-1{)}^2+(y-2{)}^2\right)=(x+y$ $+3{)}^2$ is

• A. $\left(-\frac{1}{2},-\frac{1}{2}\right)$
• B. $\left(-\frac{1}{2},\frac{1}{2}\right)$
• C. $\left(\frac{1}{2},\frac{1}{2}\right)$
• D. $\left(\frac{1}{2},-\frac{1}{2}\right)$

Answer 1

Answer: (B)

$2\left((x-1{)}^2+(y-2{)}^2\right)=(x+y+3{)}^2$
$\Rightarrow \sqrt{(x-1{)}^2+(y-2{)}^2}=\frac{|x+y+3|}{\sqrt{2}}$
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 $AS$ which is $\left(-\frac{1}{2},\frac{1}{2}\right)$.