Find normal equation

9. \begin{figure}[h] \end{figure} The curve shown in Figure 5 has equation $$x = 4 \sin ^ { 2 } y - 1 \quad 0 \leqslant y \leqslant \frac { \pi } { 2 }$$ The point \(P \left( k , \frac { \pi } { 3 } \right)\) lies on the curve.

1. Verify that \(k = 2\)
2. 1. Find \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of \(y\)
   2. Hence show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 \sqrt { x + 1 } \sqrt { 3 - x } }\) The normal to the curve at \(P\) cuts the \(x\)-axis at the point \(N\).
3. Find the exact area of triangle \(O P N\), where \(O\) is the origin. Give your answer in the form \(a \pi + b \pi ^ { 2 }\) where \(a\) and \(b\) are constants.

4 A curve has parametric equations $$x = 2 + t ^ { 2 } , \quad y = 4 t$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Find the equation of the normal at the point where \(t = 4\), giving your answer in the form \(y = m x + c\).
3. Find a cartesian equation of the curve.

1 A curve is defined by the parametric equations $$x = 1 + 2 t , \quad y = 1 - 4 t ^ { 2 }$$

1. 1. Find \(\frac { \mathrm { d } x } { \mathrm {~d} t }\) and \(\frac { \mathrm { d } y } { \mathrm {~d} t }\).
      (2 marks)
   2. Hence find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Find an equation of the normal to the curve at the point where \(t = 1\).
3. Find a cartesian equation of the curve.

2 A curve is defined, for \(t \neq 0\), by the parametric equations $$x = 4 t + 3 , \quad y = \frac { 1 } { 2 t } - 1$$ At the point \(P\) on the curve, \(t = \frac { 1 } { 2 }\).

1. Find the gradient of the curve at the point \(P\).
2. Find an equation of the normal to the curve at the point \(P\).
3. Find a cartesian equation of the curve.

2 A curve is defined by the parametric equations $$x = \frac { 1 } { t } , \quad y = t + \frac { 1 } { 2 t }$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Find an equation of the normal to the curve at the point where \(t = 1\).
3. Show that the cartesian equation of the curve can be written in the form $$x ^ { 2 } - 2 x y + k = 0$$ where \(k\) is an integer.

A curve is given parametrically by the equations $$x = 5 \cos t, \quad y = -2 + 4 \sin t, \quad 0 \leq t < 2\pi$$

1. Find the coordinates of all the points at which \(C\) intersects the coordinate axes, giving your answers in surd form where appropriate. [4]
2. Sketch the graph at \(C\). [2]

\(P\) is the point on \(C\) where \(t = \frac{1}{2}\pi\).

1. Show that the normal to \(C\) at \(P\) has equation $$8\sqrt{3}y = 10x - 25\sqrt{3}.$$ [4]

A curve is given parametrically by the equations $$x = 5 \cos t, \quad y = -2 + 4 \sin t, \quad 0 \leq t < 2\pi.$$

1. Find the coordinates of all the points at which \(C\) intersects the coordinate axes, giving your answers in surd form where appropriate. [4]
2. Sketch the graph at \(C\). [2]

\(P\) is the point on \(C\) where \(t = \frac{1}{6}\pi\).

1. Show that the normal to \(C\) at \(P\) has equation $$8\sqrt{3}y = 10x - 25\sqrt{3}.$$ [4]

A curve has parametric equations $$x = t^3 + 1, \quad y = \frac{2}{t}, \quad t \neq 0.$$

1. Find an equation for the normal to the curve at the point where \(t = 1\), giving your answer in the form \(y = mx + c\). [6]
2. Find a cartesian equation for the curve in the form \(y = f(x)\). [3]

The curve \(C\) has parametric equations $$x = 2\cos t, \quad y = \sqrt{3}\cos 2t, \quad 0 \leq t \leq \pi$$

1. Find an expression for \(\frac{dy}{dx}\) in terms of \(t\). [2]

The point \(P\) lies on \(C\) where \(t = \frac{2\pi}{3}\) The line \(l\) is the normal to \(C\) at \(P\).

1. Show that an equation for \(l\) is $$2x - 2\sqrt{3}y - 1 = 0$$ [5]

The line \(l\) intersects the curve \(C\) again at the point \(Q\).

1. Find the exact coordinates of \(Q\). You must show clearly how you obtained your answers. [6]