5.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{12d3f92f-8464-4ba1-93a2-c7b841e3d3de-4_739_1397_187_335}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{figure}
Figure 2 shows a sketch of the curve \(C\) with equation \(y = \frac { x ^ { 2 } - 2 } { x ^ { 2 } - 4 }\) and \(x \neq \pm 2\).
The curve cuts the \(y\)-axis at \(U\).
- Write down the coordinates of the point \(U\).
The point \(P\) with \(x\)-coordinate \(a ( a \neq 0 )\) lies on \(C\).
- Show that the normal to \(C\) at \(P\) cuts the \(y\)-axis at the point
$$\left( 0 , \left[ \frac { a ^ { 2 } - 2 } { a ^ { 2 } - 4 } - \frac { \left( a ^ { 2 } - 4 \right) ^ { 2 } } { 4 } \right] \right)$$
The circle \(E\), with centre on the \(y\)-axis, touches all three branches of \(C\).
- Show that
$$\left[ \frac { a ^ { 2 } } { 2 \left( a ^ { 2 } - 4 \right) } - \frac { \left( a ^ { 2 } - 4 \right) ^ { 2 } } { 4 } \right] ^ { 2 } = a ^ { 2 } + \frac { \left( a ^ { 2 } - 4 \right) ^ { 4 } } { 16 }$$
- Hence, show that
$$\left( a ^ { 2 } - 4 \right) ^ { 2 } = 1$$
- Find the centre and radius of \(E\).