Convert to Cartesian (trigonometric)

3. A curve \(C\) has parametric equations $$x = \sqrt { 3 } \tan \theta , \quad y = \sec ^ { 2 } \theta , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 3 }$$ The cartesian equation of \(C\) is $$y = \mathrm { f } ( x ) , \quad 0 \leqslant x \leqslant k , \quad \text { where } k \text { is a constant }$$

1. State the value of \(k\).
2. Find \(\mathrm { f } ( x )\) in its simplest form.
3. Hence, or otherwise, find the gradient of the curve at the point where \(\theta = \frac { \pi } { 6 }\)

3. The curve \(C\) has parametric equations $$x = 3 + 2 \sin t \quad y = \frac { 6 } { 7 + \cos 2 t } \quad - \frac { \pi } { 2 } \leqslant t \leqslant \frac { \pi } { 2 }$$

1. Show that \(C\) has Cartesian equation $$y = \frac { 12 } { ( 7 - x ) ( 1 + x ) } \quad p \leqslant x \leqslant q$$ where \(p\) and \(q\) are constants to be found.
2. Hence, find a Cartesian equation for \(C\) in the form $$y = \frac { a } { x + b } + \frac { c } { x + d } \quad p \leqslant x \leqslant q$$ where \(a , b , c\) and \(d\) are constants.

1. A curve has parametric equations

$$x = 2 \cot t , \quad y = 2 \sin ^ { 2 } t , \quad 0 < t \leqslant \frac { \pi } { 2 }$$

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of the parameter \(t\).
2. Find an equation of the tangent to the curve at the point where \(t = \frac { \pi } { 4 }\).
3. Find a cartesian equation of the curve in the form \(y = \mathrm { f } ( x )\). State the domain on which the curve is defined.

6. A curve has parametric equations $$x = \tan ^ { 2 } t , \quad y = \sin t , \quad 0 < t < \frac { \pi } { 2 }$$

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\). You need not simplify your answer.
2. Find an equation of the tangent to the curve at the point where \(t = \frac { \pi } { 4 }\). Give your answer in the form \(y = a x + b\), where \(a\) and \(b\) are constants to be determined.
3. Find a cartesian equation of the curve in the form \(y ^ { 2 } = \mathrm { f } ( x )\).
   \section*{LU}

5. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve with parametric equations $$x = 2 \cos 2 t , \quad y = 6 \sin t , \quad 0 \leqslant t \leqslant \frac { \pi } { 2 }$$

1. Find the gradient of the curve at the point where \(t = \frac { \pi } { 3 }\).
2. Find a cartesian equation of the curve in the form $$y = \mathrm { f } ( x ) , \quad - k \leqslant x \leqslant k$$ stating the value of the constant \(k\).
3. Write down the range of \(\mathrm { f } ( x )\).

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

$$x = 2 \sin t , \quad y = 1 - \cos 2 t , \quad - \frac { \pi } { 2 } \leqslant t \leqslant \frac { \pi } { 2 }$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at the point where \(t = \frac { \pi } { 6 }\)
2. Find a cartesian equation for \(C\) in the form $$y = \mathrm { f } ( x ) , \quad - k \leqslant x \leqslant k$$ stating the value of the constant \(k\).
3. Write down the range of \(\mathrm { f } ( x )\).

5. \begin{figure}[h] \end{figure} The curve shown in Fig. 1 has parametric equations $$x = \cos t , y = \sin 2 t , \quad 0 \leq t < 2 \pi .$$

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of the parameter \(t\).
2. Find the values of the parameter \(t\) at the points where \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\).
3. Hence give the exact values of the coordinates of the points on the curve where the tangents are parallel to the \(x\)-axis.
4. Show that a cartesian equation for the part of the curve where \(0 \leq t < \pi\) is $$y = 2 x \sqrt { } \left( 1 - x ^ { 2 } \right)$$
5. Write down a cartesian equation for the part of the curve where \(\pi \leq t < 2 \pi\).
   5. continued

5. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve \(C\) with parametric equations $$x = 5 + 2 \tan t \quad y = 8 \sec ^ { 2 } t \quad - \frac { \pi } { 3 } \leqslant t \leqslant \frac { \pi } { 4 }$$

1. Use parametric differentiation to find the gradient of \(C\) at \(x = 3\) The curve \(C\) has equation \(y = \mathrm { f } ( x )\), where f is a quadratic function.
2. Find \(\mathrm { f } ( x )\) in the form \(a ( x + b ) ^ { 2 } + c\), where \(a\), \(b\) and \(c\) are constants to be found.
3. Find the range of f.

7. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of the curve \(C\) with parametric equations $$x = 4 \cos \left( t + \frac { \pi } { 6 } \right) \quad y = 2 \sin t \quad 0 \leqslant t \leqslant 2 \pi$$

1. Show that $$x + y = 2 \sqrt { 3 } \cos t$$
2. Show that a cartesian equation of \(C\) is $$( x + y ) ^ { 2 } + a y ^ { 2 } = b$$ where \(a\) and \(b\) are integers to be found. \includegraphics[max width=\textwidth, alt={}, center]{4de08317-5fb9-4789-8d57-ccf463224c78-22_2673_1948_107_118}

5 A curve \(C\) has parametric equations $$x = \cos t , \quad y = 3 + 2 \cos 2 t , \quad \text { where } 0 \leqslant t \leqslant \pi$$

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\) and hence show that the gradient at any point on \(C\) cannot exceed 8 .
2. Show that all points on \(C\) satisfy the cartesian equation \(y = 4 x ^ { 2 } + 1\).
3. Sketch the curve \(y = 4 x ^ { 2 } + 1\) and indicate on your sketch the part which represents \(C\).

3 The parametric equations of a curve are $$x = \cos 2 \theta , \quad y = \sin \theta \cos \theta \quad \text { for } 0 \leqslant \theta < \pi$$ Show that the cartesian equation of the curve is \(x ^ { 2 } + 4 y ^ { 2 } = 1\).
Sketch the curve.

9 Two astronomers wish to model the path of motion of a particle in two dimensions.
Experimental results show that the position of the particle can be found using the parametric equations $$x = 2 \cos \theta - \sin \theta + 2 \quad y = \cos \theta + 2 \sin \theta - 1 \quad \left( 0 \leq \theta \leq 360 ^ { \circ } \right)$$ One astronomer uses trigonometry.

1. Express \(2 \cos \theta - \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R\) and \(\alpha\) are constants to be determined. Show also that, for the same values of \(R\) and \(\alpha\), $$\cos \theta + 2 \sin \theta = R \sin ( \theta + \alpha )$$
2. Hence, or otherwise, show that the path of particle may be written in the form $$( x - 2 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 5$$ Describe the path of the particle. The second astronomer sets up a first order differential equation with the condition that \(x = 4\) when \(y = 0\).
3. Verify that the point with parameter \(\theta = 0\) has coordinates \(( 4,0 )\).
4. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Deduce that \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { x - 2 } { y + 1 }$$
5. Solve this differential equation, using the condition that \(y = 0\) when \(x = 4\). Hence show that the two solutions give the same cartesian equation for the path of particle.

4. \includegraphics[max width=\textwidth, alt={}, center]{c7b867af-0730-459e-9c76-15eb07b9e476-1_465_976_1539_388} The diagram shows the curve with parametric equations $$x = \tan \theta , \quad y = \cos ^ { 2 } \theta , \quad - \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 }$$

1. Find a cartesian equation for the curve. The shaded region is bounded by the curve, the \(x\)-axis and the lines \(x = - 1\) and \(x = 1\).
2. Using integration, with the substitution \(x = \tan u\), find the area of the shaded region.

9. A curve has parametric equations $$x = \sec \theta + \tan \theta , \quad y = \operatorname { cosec } \theta + \cot \theta , \quad 0 < \theta < \frac { \pi } { 2 }$$

1. Show that \(x + \frac { 1 } { x } = 2 \sec \theta\). Given that \(y + \frac { 1 } { y } = 2 \operatorname { cosec } \theta\),
2. find a cartesian equation for the curve.
3. Show that \(\frac { \mathrm { d } x } { \mathrm {~d} \theta } = \frac { 1 } { 2 } \left( x ^ { 2 } + 1 \right)\).
4. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).

4 The parametric equations of a curve are $$x = \cos 2 \theta , \quad y = \sin \theta \cos \theta \quad \text { for } 0 \leqslant \theta < \pi$$ Show that the cartesian equation of the curve is \(x ^ { 2 } + 4 y ^ { 2 } = 1\).
Sketch the curve.

1 A curve has parametric equations \(x = \sec \theta , y = 2 \tan \theta\).

1. Given that the derivative of \(\sec \theta\) is \(\sec \theta \tan \theta\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \operatorname { cosec } \theta\).
2. Verify that the cartesian equation of the curve is \(y ^ { 2 } = 4 x ^ { 2 } - 4\). Fig. 5 shows the region enclosed by the curve and the line \(x = 2\). This region is rotated through \(180 ^ { \circ }\) about the \(x\)-axis. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1e788cb0-36b0-42a9-9e0c-077022d410ae-1_556_867_580_588} \captionsetup{labelformat=empty} \caption{Fig. 5}
   \end{figure}
3. Find the volume of revolution produced, giving your answer in exact form.

8 A curve is defined by the parametric equations $$x = \sin ^ { 2 } \theta , \quad y = 4 \sin \theta - \sin ^ { 3 } \theta ,$$ where \(- \frac { 1 } { 2 } \pi \leqslant \theta \leqslant \frac { 1 } { 2 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 - 3 \sin ^ { 2 } \theta } { 2 \sin \theta }\).
2. Find the coordinates of the point on the curve at which the gradient is 2 .
3. Show that the curve has no stationary points.
4. Find a cartesian equation of the curve, giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\).

7 A curve has parametric equations $$x = 2 \sin t , \quad y = \cos 2 t + 2 \sin t$$ for \(- \frac { 1 } { 2 } \pi \leqslant t \leqslant \frac { 1 } { 2 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1 - 2 \sin t\) and hence find the coordinates of the stationary point.
2. Find the cartesian equation of the curve.
3. State the set of values that \(x\) can take and hence sketch the curve.

3 The parametric equations of a curve are $$x = \sin \theta , \quad y = \sin 2 \theta , \quad \text { for } 0 \leqslant \theta \leqslant 2 \pi .$$

1. Find the exact value of the gradient of the curve at the point where \(\theta = \frac { 1 } { 6 } \pi\).
2. Show that the cartesian equation of the curve is \(y ^ { 2 } = 4 x ^ { 2 } - 4 x ^ { 4 }\).

5 A curve has parametric equations \(x = \sec \theta , y = 2 \tan \theta\).

1. Given that the derivative of \(\sec \theta\) is \(\sec \theta \tan \theta\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \operatorname { cosec } \theta\).
2. Verify that the cartesian equation of the curve is \(y ^ { 2 } = 4 x ^ { 2 } - 4\). Fig. 5 shows the region enclosed by the curve and the line \(x = 2\). This region is rotated through \(180 ^ { \circ }\) about the \(x\)-axis. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{132ae754-bd4c-4819-80ef-4823ac2ead4f-02_545_853_1738_607} \captionsetup{labelformat=empty} \caption{Fig. 5}
   \end{figure}
3. Find the volume of revolution produced, giving your answer in exact form.

14. A curve \(C\) has parametric equations $$x = 1 - \cos t , \quad y = 2 \cos 2 t , \quad 0 \leq t < \pi$$ a. Show that the cartesian equation of the curve can be written as \(y = k ( 1 - x ) ^ { 2 } - 2\) where \(k\) is an integer.
b. i. Sketch the curve C .
ii. Explain briefly why C does not include all points of \(y = k ( 1 - x ) ^ { 2 } - 2 , x \in \mathbb { R }\). The line with equation \(y = k - x\), where \(k\) is a constant, intersects C at two distinct points.
(c) State the range of values of \(k\), writing your answer in set notation.

16. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of the curve \(C\) with parametric equations \(x = - 3 + 6 \sin \theta , \quad y = 9 \cos 2 \theta \quad - \frac { \pi } { 2 } \leq \theta \leq \frac { \pi } { 4 }\) where \(\theta\) is a parameter.
a. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\) The line \(l\) is normal to \(C\) at the point \(P\) where \(\theta = \frac { \pi } { 6 }\) b. Show that an equation for \(l\) is $$y = \frac { 1 } { 3 } x + \frac { 9 } { 2 }$$ c. The cartesian equation for the curve \(C\) can be written in the form $$y = a - \frac { 1 } { 2 } ( x + b ) ^ { 2 }$$ where \(a\) and \(b\) are integers to be found. The straight line with equation $$y = \frac { 1 } { 3 } x + k$$ where \(k\) is a constant intersects \(C\) at two distinct points.
d. Find the range of possible values for \(k\).

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

$$x = 3 + 2 \sin t , \quad y = 4 + 2 \cos 2 t , \quad 0 \leqslant t < 2 \pi$$

1. Show that all points on \(C\) satisfy \(y = 6 - ( x - 3 ) ^ { 2 }\)
2. 1. Sketch the curve \(C\).
   2. Explain briefly why \(C\) does not include all points of \(y = 6 - ( x - 3 ) ^ { 2 } , \quad x \in \mathbb { R }\) The line with equation \(x + y = k\), where \(k\) is a constant, intersects \(C\) at two distinct points.
3. State the range of values of \(k\), writing your answer in set notation.

14. \begin{figure}[h] \end{figure} Figure 6 shows a sketch of the curve \(C\) with parametric equations $$x = 4 \cos \left( t + \frac { \pi } { 6 } \right) , y = 2 \sin t , \quad 0 < t \leqslant 2 \pi$$ Show that a Cartesian equation of \(C\) can be written in the form $$( x + y ) ^ { 2 } + a y ^ { 2 } = b$$ where \(a\) and \(b\) are integers to be found.

16. \begin{figure}[h] \end{figure} Figure 6 shows a sketch of the curve \(C\) with parametric equations $$x = 2 \tan t + 1 \quad y = 2 \sec ^ { 2 } t + 3 \quad - \frac { \pi } { 4 } \leqslant t \leqslant \frac { \pi } { 3 }$$ The line \(l\) is the normal to \(C\) at the point \(P\) where \(t = \frac { \pi } { 4 }\)

1. Using parametric differentiation, show that an equation for \(l\) is $$y = - \frac { 1 } { 2 } x + \frac { 17 } { 2 }$$
2. Show that all points on \(C\) satisfy the equation $$y = \frac { 1 } { 2 } ( x - 1 ) ^ { 2 } + 5$$ The straight line with equation $$y = - \frac { 1 } { 2 } x + k \quad \text { where } k \text { is a constant }$$ intersects \(C\) at two distinct points.
3. Find the range of possible values for \(k\).