Question

If origin is shifted to the point (1,-2) the equation $y^2-4y+8=0$ becomes:

Answer 1

Assuming a correction in the equation as
$y^2-4y+8=x$
We can solve the question as follows:
when origin is shifted to (1,-2), all the points in the old plane will be shifted and points in the new plane can be defined as
x(new)= x(old) +1
y(new)= y(old) - 2
so transforming the older equation $y^2-4y+8=x$, we can write
$(y-2{)}^2-4(y-2)+8=x+1$
therefore, it simplifies as $y^2-8y+19=x$
The roots to this equation will be imaginary since $(b^2-4ac)<0$.