Answer
$(y'+1)^2=-(x'-3)$, vertex $(3,-1)$
Work Step by Step
Step 1. Given
$x^{2}-4xy+4y^{2}+5\sqrt 5y-10=0$
we have
$A=1, B=-4, C=4$
and the rotation angle is
$cot(2\theta)=\frac{A-C}{B}=\frac{1-4}{-4}=\frac{3}{4}$
Step 2. We have the transformations (see also exercise 15):
$x=\frac{\sqrt 5}{5}(2x'-y')$ and $y=\frac{\sqrt 5}{5}(x'+2y')$
Step 3. The equation becomes
$x^{2}-4xy+4y^{2}+5\sqrt 5y-10=(\frac{\sqrt 5}{5}(2x'-y'))^{2}-4(\frac{\sqrt 5}{5}(2x'-y'))(\frac{\sqrt 5}{5}(x'+2y'))+4(\frac{\sqrt 5}{5}(x'+2y'))^{2}+5\sqrt 5(\frac{\sqrt 5}{5}(x'+2y'))-10=5x;+5y'^2+10y'-10$
or, $x'+y'^2+2y'-2=0$ and $x'+(y'+1)^2-3=0$
Step 4. In standard form, we have $(y'+1)^2=-(x'-3)$, which represents a parabola.
Step 5. We can find the vertex in the $x'-y'$ system as $(3,-1)$
