Convert to Cartesian (polynomial/rational)

2. A curve \(C\) has parametric equations $$x = \frac { 3 } { 2 } t - 5 , \quad y = 4 - \frac { 6 } { t } \quad t \neq 0$$

1. Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at \(t = 3\), giving your answer as a fraction in its simplest form.
2. Show that a cartesian equation of \(C\) can be expressed in the form $$y = \frac { a x + b } { x + 5 } \quad x \neq k$$ where \(a , b\) and \(k\) are integers to be found.

10. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve \(C\) with parametric equations $$x = \frac { 20 t } { 2 t + 1 } \quad y = t ( t - 4 ) , \quad t > 0$$ The curve cuts the \(x\)-axis at the point \(P\).

1. Find the \(x\) coordinate of \(P\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( t - A ) ( 2 t + 1 ) ^ { 2 } } { B }\) where \(A\) and \(B\) are constants to be found.
3. 1. Make \(t\) the subject of the formula $$x = \frac { 20 t } { 2 t + 1 }$$
   2. Hence find a cartesian equation of the curve \(C\). Write your answer in the form $$y = \mathrm { f } ( x ) , \quad 0 < x < k$$ where \(\mathrm { f } ( x )\) is a single fraction and \(k\) is a constant to be found.

4. The curve \(C\) is defined by the parametric equations $$x = \frac { 1 } { t } + 2 \quad y = \frac { 1 - 2 t } { 3 + t } \quad t > 0$$

1. Show that the equation of \(C\) can be written in the form \(y = \mathrm { g } ( x )\) where g is the function $$\mathrm { g } ( x ) = \frac { a x + b } { c x + d } \quad x > k$$ where \(a , b , c , d\) and \(k\) are integers to be found.
2. Hence, or otherwise, state the range of g .
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1. A curve \(C\) has parametric equations

$$x = 2 t + 5 , \quad y = 3 + \frac { 4 } { t } , \quad t \neq 0$$

1. Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at the point on \(C\) with coordinates \(( 9,5 )\).
2. Find a cartesian equation of the curve in the form $$y = \frac { a x + b } { c x + d }$$ where \(a\), \(b\), \(c\) and \(d\) are integers.

1. The curve \(C\) has parametric equations

$$x = 3 t - 4 , y = 5 - \frac { 6 } { t } , \quad t > 0$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\) The point \(P\) lies on \(C\) where \(t = \frac { 1 } { 2 }\)
2. Find the equation of the tangent to \(C\) at the point \(P\). Give your answer in the form \(y = p x + q\), where \(p\) and \(q\) are integers to be determined.
3. Show that the cartesian equation for \(C\) can be written in the form $$y = \frac { a x + b } { x + 4 } , \quad x > - 4$$ where \(a\) and \(b\) are integers to be determined.

1. A curve \(C\) has parametric equations

$$x = \frac { t } { t - 3 } \quad y = \frac { 1 } { t } + 2 \quad t \in \mathbb { R } \quad t > 3$$ Show that all points on \(C\) lie on the curve with Cartesian equation $$y = \frac { a x - 1 } { b x }$$ where \(a\) and \(b\) are constants to be found.

4 A curve is defined by parametric equations $$x = \frac { 1 } { t } - 1 , y = \frac { 2 + t } { 1 + t }$$ Show that the cartesian equation of the curve is \(y = \frac { 3 + 2 x } { 2 + x }\).

4 Find the first three terms in the binomial expansion of \(\sqrt { 4 + x }\) in ascending powers of \(x\).
State the set of values of \(x\) for which the expansion is valid.
show that \(\frac { y - 2 } { y + 1 } = A \mathrm { e } ^ { x ^ { 3 } }\), where \(A\) is a constant.
(ii) Hence, given that \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { 2 } ( y - 2 ) ( y + 1 ) ,$$ 6 Solve the equation \(\tan \left( \theta + 45 ^ { \circ } \right) = 1 - 2 \tan \theta\), for \(0 ^ { \circ } \leqslant \theta \leqslant 90 ^ { \circ }\). Section B (36 marks)
7 A straight pipeline AB passes through a mountain. With respect to axes \(\mathrm { O } x y z\), with \(\mathrm { O } x\) due East, \(\mathrm { O } y\) due North and \(\mathrm { O } z\) vertically upwards, A has coordinates \(( - 200,100,0 )\) and B has coordinates \(( 100,200,100 )\), where units are metres.

5 A ball is thrown towards a hedge. Its position relative to the point from which it was thrown is given by the parametric equations $$x = 8 t , y = 10 t - 5 t ^ { 2 }$$

1. Find the cartesian equation of the trajectory of the ball.
2. The ball just clears the hedge. What can you say about the height of the hedge?

6. A curve has parametric equations $$x = \frac { t } { 2 - t } , \quad y = \frac { 1 } { 1 + t } , \quad - 1 < t < 2$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 1 } { 2 } \left( \frac { 2 - t } { 1 + t } \right) ^ { 2 }\).
2. Find an equation for 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 $$y = \frac { 1 + x } { 1 + 3 x }$$

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 \(y = m x + c\).
2. Find a cartesian equation for the curve in the form \(y = \mathrm { f } ( x )\).

8 A curve is defined by parametric equations $$x = \frac { 1 } { t } - 1 , y = \frac { 2 + t } { 1 + t }$$ Show that the cartesian equation of the curve is \(y = \frac { 3 + 2 x } { 2 + x }\).

5 A curve is defined parametrically by the equations $$x = \frac { 1 } { 1 + t } , \quad y = \frac { 1 - t } { 1 + 2 t }$$ Find \(t\) in terms of \(x\). Hence find the cartesian equation of the curve, giving your answer as simply as possible.

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.

5 A curve has parametric equations $$x = 2 t + t ^ { 2 } , \quad y = 2 t ^ { 2 } + t ^ { 3 }$$

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\) and find the gradient of the curve at the point \(( 3 , - 9 )\).
2. By considering \(\frac { y } { x }\), find a cartesian equation of the curve, giving your answer in a form not involving fractions.

7 The parametric equations of a curve are \(x = \frac { t + 2 } { t + 1 } , y = \frac { 2 } { t + 3 }\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } > 0\).
2. Find the cartesian equation of the curve, giving your answer in a form not involving fractions.

2 A curve is defined parametrically by the equations $$x = \frac { 1 } { 1 + t } , \quad y = \frac { 1 - t } { 1 + 2 t }$$ Find \(t\) in terms of \(x\). Hence find the cartesian equation of the curve, giving your answer as simply as possible.

10 A curve has parametric equations \(x = t + \frac { 2 } { t }\) and \(y = t - \frac { 2 } { t }\), for \(t \neq 0\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\), giving your answer in its simplest form.
2. Explain why the curve has no stationary points.
3. By considering \(x + y\), or otherwise, find a cartesian equation of the curve, giving your answer in a form not involving fractions or brackets.

5. A curve \(C\) has parametric equations $$x = 2 t - 1 , \quad y = 4 t - 7 + \frac { 3 } { t } , \quad t \neq 0$$ Show that the Cartesian equation of the curve \(C\) can be written in the form $$y = \frac { 2 x ^ { 2 } + a x + b } { x + 1 } , \quad x \neq - 1$$ where \(a\) and \(b\) are integers to be found.

12 Fig. 12 shows a curve C with parametric equations \(x = 4 t ^ { 2 } , y = 4 t\). The point P , with parameter \(t\), is a general point on the curve. Q is the point on the line \(x + 4 = 0\) such that PQ is parallel to the \(x\)-axis. R is the point \(( 4,0 )\). \begin{figure}[h] \end{figure}

1. Show algebraically that P is equidistant from Q and R .
2. Find a cartesian equation of C .

8 A particle moves in the \(x - y\) plane so that its position at time \(t\) s is given by \(x = t ^ { 3 } - 8 t , y = t ^ { 2 }\) for \(- 3.5 < t < 3.5\). The units of distance are metres. The graph shows the path of the particle and the direction of travel at the point \(\mathrm { P } ( 8,4 )\).
\includegraphics[max width=\textwidth, alt={}, center]{9dd6fc6d-b51e-4a73-ace5-d26a7558032c-07_492_924_415_242}

1. Find \(\frac { \mathrm { dy } } { \mathrm { dx } }\) in terms of \(t\).
2. Hence show that the value of \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at P is - 1 .
3. Find the time at which the particle is travelling in the direction opposite to that at P .
4. Find the cartesian equation of the path, giving \(x ^ { 2 }\) as a function of \(y\).

5 A curve is defined by the parametric equations $$x = 8 t ^ { 2 } - t , \quad y = \frac { 3 } { t }$$

1. Show that the cartesian equation of the curve can be written as \(x y ^ { 2 } + 3 y = k\), stating the value of the integer \(k\).
   (2 marks)
2. 1. Find an equation of the tangent to the curve at the point \(P\), where \(t = \frac { 1 } { 4 }\).
   2. Verify that the tangent at \(P\) intersects the curve when \(x = \frac { 3 } { 2 }\).

2 A curve is defined by the parametric equations $$x = 1 - 3 t , \quad y = 1 + 2 t ^ { 3 }$$

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. Find a cartesian equation of the curve.

1 A curve is defined by the parametric equations \(x = \frac { t ^ { 2 } } { 2 } + 1 , y = \frac { 4 } { t } - 1\).

1. Find the gradient at the point on the curve where \(t = 2\).
2. Find a Cartesian equation of the curve.
   \includegraphics[max width=\textwidth, alt={}]{9f03a5f3-7fea-4fb7-b3bd-b4c0cdf662a2-02_1730_1709_977_153}

5. A curve has parametric equations $$x = \frac { t } { 2 - t } , \quad y = \frac { 1 } { 1 + t } , \quad - 1 < t < 2$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 1 } { 2 } \left( \frac { 2 - t } { 1 + t } \right) ^ { 2 }\).
2. Find an equation for 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 $$y = \frac { 1 + x } { 1 + 3 x }$$
   1. continued
   2. (a) Find \(\int \tan ^ { 2 } x d x\).
   3. Show that
   $$\int \tan x \mathrm {~d} x = \ln | \sec x | + c$$ where \(c\) is an arbitrary constant. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{fe01157a-7617-43d3-900c-8d043bcbe784-10_566_789_648_504} \captionsetup{labelformat=empty} \caption{Figure 1}
   \end{figure} Figure 1 shows part of the curve with equation \(y = x ^ { \frac { 1 } { 2 } } \tan x\).
   The shaded region bounded by the curve, the \(x\)-axis and the line \(x = \frac { \pi } { 3 }\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
4. Show that the volume of the solid formed is \(\frac { 1 } { 18 } \pi ^ { 2 } ( 6 \sqrt { 3 } - \pi ) - \pi \ln 2\).