Problem source

The transformation $T$ from $\begin{pmatrix} x \\ y \end{pmatrix}$ to $\begin{pmatrix} X \\ Y \end{pmatrix}$ is given by 
\[
\begin{pmatrix}X\\
Y
\end{pmatrix}=\frac{2}{5}\begin{pmatrix}9 & -2\\
-2 & 6
\end{pmatrix}\begin{pmatrix}x\\
y
\end{pmatrix}.
\]
Show that $T$ leaves the vector $\begin{pmatrix} 1\\ 2 \end{pmatrix}$ unchanged in direction but multiplied by a scalar, and that $\begin{pmatrix} 2\\ -1 \end{pmatrix}$ is similarly transformed. 
The circle $C$ whose equation is $x^{2}+y^{2}=1$ transforms under $T$ to a curve $E$. Show that $E$ has equation 
\[
8X^{2}+12XY+17Y^{2}=80,
\]
and state the area of the region bounded by $E$. Show also that the greatest value of $X$ on $E$ is $2\sqrt{17/5}.$ 
Find the equation of the tangent to $E$ at the point which corresponds to the point $\frac{1}{5}(3,4)$ on $C$.

Solution source

\begin{align*}
T\begin{pmatrix}1\\
2
\end{pmatrix} &= \frac{2}{5}\begin{pmatrix}9 & -2\\
-2 & 6
\end{pmatrix}\begin{pmatrix}1\\
2
\end{pmatrix} \\
&= \frac25\begin{pmatrix}9 - 4\\
-2+12
\end{pmatrix} \\
&= \begin{pmatrix}2\\
4
\end{pmatrix} \\
&= 2 \begin{pmatrix}1\\
2
\end{pmatrix}
\end{align*}

\begin{align*}
T\begin{pmatrix}1\\
2
\end{pmatrix} &= \frac{2}{5}\begin{pmatrix}9 & -2\\
-2 & 6
\end{pmatrix}\begin{pmatrix}2\\
-1
\end{pmatrix} \\
&= \frac25\begin{pmatrix}18+2\\
-4-6
\end{pmatrix} \\
&= \begin{pmatrix}8\\
-4
\end{pmatrix} \\
&= 4 \begin{pmatrix}2\\
-1
\end{pmatrix}
\end{align*}

Consider $T^{-1} = \frac{5}{2} \frac{1}{50}\begin{pmatrix}6 & 2\\
2 & 9
\end{pmatrix}$, so $T^{-1} \begin{pmatrix}X\\
Y
\end{pmatrix} = \begin{pmatrix}x\\
y
\end{pmatrix}$ and so:

\begin{align*}
x^2 + y^2 & = \begin{pmatrix}x&
y
\end{pmatrix}\begin{pmatrix}x\\
y
\end{pmatrix} \\
&= \begin{pmatrix}X&
Y
\end{pmatrix} (T^{-1})^T T^{-1} \begin{pmatrix}X\\
Y
\end{pmatrix} \\
&= \frac{1}{400}\begin{pmatrix}X&
Y
\end{pmatrix}\begin{pmatrix}6 & 2\\
2 & 9
\end{pmatrix}\begin{pmatrix}6 & 2\\
2 & 9
\end{pmatrix} \begin{pmatrix}X\\
Y
\end{pmatrix} \\
&= \frac{1}{400}\begin{pmatrix}X&
Y
\end{pmatrix}\begin{pmatrix}6 & 2\\
2 & 9
\end{pmatrix} \begin{pmatrix}6X+2Y\\
2X+9Y
\end{pmatrix} \\
&= \frac{1}{400}\begin{pmatrix}X&
Y
\end{pmatrix} \begin{pmatrix}6(6X+2Y)+2(2X+9Y)\\
2(6X+2Y)+9(2X+9Y)
\end{pmatrix} \\
&= \frac{1}{400}\begin{pmatrix}X&
Y
\end{pmatrix} \begin{pmatrix}40X+30Y\\
30X +85Y
\end{pmatrix} \\
&= \frac{1}{80}\begin{pmatrix}X&
Y
\end{pmatrix} \begin{pmatrix}8X+6Y\\
6X +17Y
\end{pmatrix} \\
&= \frac{1}{80} \l 8X^2 + 12XY + 17Y^2\r
\end{align*}

Therefore $8X^2 + 12XY + 17Y^2 = 80$. The area will be $\det T \cdot \pi = \frac{4}{25} \cdot 50 \cdot \pi = 8 \pi$.

Differentiating we obtain $2 \cdot 8 \cdot X \cdot \frac{dX}{dY} + 2 \cdot 6 \cdot X + 2 \cdot 6 \cdot Y \cdot \frac{dX}{dY} + 2 \cdot 17 Y \Rightarrow \frac{dX}{dY} = -\frac{6X + 17Y}{8X+6Y}$, at a maximum (or minimum, $6X = -17Y$). Therefore
\begin{align*} 
\Rightarrow && 8X^2 + 12 \cdot \frac{6}{17}X^2 + 17 ( -\frac{6}{17} X)^2 &= 80 \\
\Rightarrow && \frac{100}{17}X^2 &= 80 \\
\Rightarrow &&X^2 &= \frac{17 \cdot 4}{5} \\
\Rightarrow && |X| = 2 \sqrt {\frac{17}{5}}
\end{align*}

The point $\frac15 (3,4)$ maps to 

\begin{align*}
\frac{2}{5}\frac{1}{5}\begin{pmatrix}9 & -2\\
-2 & 6
\end{pmatrix}\begin{pmatrix}3\\
4
\end{pmatrix} &= \frac{2}{25} \begin{pmatrix}19\\
18
\end{pmatrix}
\end{align*}

So the point is $(\frac{38}{25}, \frac{36}{25})$, with gradient $\frac{dY}{dX} =  -\frac{8X+6Y}{6X + 17Y}$ which is $-\frac{8 \cdot 19+6 \cdot 18}{6\cdot 19 + 17 \cdot 18} = -\frac{13}{21}$

therefore the equation is $21Y+13X = 50$