8 The curve defined by the parametric equations
$$x = t ^ { 2 } \text { and } y = 2 t \quad - \sqrt { 2 } \leq t \leq \sqrt { 2 }$$
is shown in Figure 1 below.
\begin{figure}[h]
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{27339c29-c4a1-480c-b882-930f8dacc7af-13_1063_1022_607_507}
\end{figure}
8
- Find a Cartesian equation of the curve in the form \(y ^ { 2 } = \mathrm { f } ( x )\)
8 - The point \(A\) lies on the curve where \(t = a\)
The tangent to the curve at \(A\) is at an angle \(\theta\) to a line through \(A\) parallel to the \(x\)-axis.
The point \(B\) has coordinates \(( 1,0 )\)
The line \(A B\) is at an angle \(\phi\) to the \(x\)-axis.
\includegraphics[max width=\textwidth, alt={}, center]{27339c29-c4a1-480c-b882-930f8dacc7af-14_846_936_678_552}
8 - By considering the gradient of the curve, show that
$$\tan \theta = \frac { 1 } { a }$$
8
- (ii) Find \(\tan \phi\) in terms of \(a\).
8
- (iii) Show that \(\tan 2 \theta = \tan \phi\)