I have the following question:
The graph has of $y=\sqrt{16-x^2}$ has been transformed into the graph of $y=5-2\sqrt{55-6x-x^2}$. Using the following matrix, I have to apply the transformation and solve for the variables:
$$ T \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} + \begin{bmatrix} h\\ k \end{bmatrix} $$
I tried to expand the matrix by getting the transformed $x$ and $y$ in terms of the old $x$ and $y$.
$x'=ax+h$
$y'=by+k$
->$x=\frac{1}{a}(x'-h)$
->$y=\frac{1}{b}(y'-k)$
->$y=\sqrt{16-x^2}$ ->$ \frac{1}{b}(y'-k)=\sqrt{16-(\frac{1}{a}(x'-h))^2}$
After expanding:
$ \frac{1}{b}(y'-k) =\sqrt{16-\frac{h^2}{a^2} +\frac{2xh}{a^2} - \frac{x'^2}{a^2}}$
If we compare this equation(only the terms under the square root sign) to the new one they gave, then:
$\sqrt{16-\frac{h^2}{a^2} +\frac{2xh}{a^2} - \frac{x'^2}{a^2}}$ = $\sqrt{55-6x-x^2}$
->$\frac{2h}{a^2} = -6$
-> $16-\frac{h^2}{a^2} = 55$
-> $- \frac{1}{a^2} = -1$
But when I try to solve for $h$ and $a$, it does not add up with the simultaneous equations I got. I think I have done the algebra correctly, I have not found an error(unless I missed it) in my arithmetic, so where am I going wrong?
Before trying to match terms, you should solve for $y'$:
$$ y' = k + b \sqrt{16 - \left(\frac{x'-h}{a}\right)^2} $$ Since we want the coefficient of the square root to be $-2$, let's define $b = -2c$ where $c > 0$. We can then insert $c$ into the square root (this is the crucial step that you missed):
\begin{align} y' &= k - 2c\sqrt{16 - \left(\frac{x'-h}{a}\right)^2} \\ &= k - 2 \sqrt{16c^2 - \left(\frac{x'-h}{a}\right)^2 c^2} \\ &= k - 2 \sqrt{16c^2 - \frac{c^2}{a^2}(x'-h)^2} \\ &= k - 2 \sqrt{16c^2 - \frac{c^2}{a^2}(x'^2 - 2hx' + h^2)} \\ &= k - 2 \sqrt{16c^2 - \frac{c^2h^2}{a^2} + \frac{2c^2h}{a^2}x' - \frac{c^2}{a^2}x'^2} \\ \end{align}
Now, you can start matching:
\begin{align} k &= 5 \\ 16c^2 - \frac{c^2h^2}{a^2} &= 55 \\ \frac{2c^2h}{a^2} &= -6 \\ \frac{c^2}{a^2} &= 1 \end{align}
These can be easily solved to give $h=-3, c = |a| = 2, b= -4$. If you choose $a=-2$ you'd be flipping the $x$ axis in the other direction, but otherwise, nothing would change.
EDIT: you can see that if you don't introduce the $b$ (or the $c$) above under the square root, the constant term under the square root is going to be 16 minus a square ($h^2/a^2$ in your equation in the question): since the square is always positive, that constant term is going to be less than 16. But then there is no way to match it against the desired constant term, which is 55, so you cannot get a solution this way. What getting the $c$ under the square root does is it multiplies that smaller quantity by 4 and gets it up to the desired 55.