Finding affine transformation

Question

Find affine transformation which takes the ellipse $x^2+4y^2+2x-8y+3=0$ to the form of the ellipse ${x^2 \over 9}+{y^2 \over 16}=1$.

So I took the quadric and reached to a standard form: ${(x+1)^2 \over 2}+{(y-1)^2 \over {1 \over 2}}=1$.

So the constant vector in the affine transformation is $(-1,1)$. But how do I find the matrix needed? [Affine transformation as I know it is $T_A(\vec v)=A \vec v+\vec a$, so I found $\vec a=(-1,1)$].

Thanks in advance for any assistance!

Answer 1

HINT : $$\frac{(x+1)^2}{2}+\frac{(y-1)^2}{\frac 12}=1$$ $$\Rightarrow \frac{x^2}{2}+\frac{y^2}{\frac 12}=1$$ $$\Rightarrow \frac{\left(\frac{\sqrt 2}{3}x\right)^2}{2}+\frac{\left(\frac{1}{4\sqrt 2}y\right)^2}{\frac 12}=1$$ $$\Rightarrow \frac{x^2}{9}+\frac{y^2}{16}=1$$