Properties of specific curves

6 The parametric equations of a curve are $$x = a \cos ^ { 3 } t , \quad y = a \sin ^ { 3 } t$$ where \(a\) is a positive constant and \(0 < t < \frac { 1 } { 2 } \pi\).

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Show that the equation of the tangent to the curve at the point with parameter \(t\) is $$x \sin t + y \cos t = a \sin t \cos t$$
3. Hence show that, if this tangent meets the \(x\)-axis at \(X\) and the \(y\)-axis at \(Y\), then the length of \(X Y\) is always equal to \(a\).

5 The parametric equations of a curve are $$x = a \cos ^ { 4 } t , \quad y = a \sin ^ { 4 } t$$ where \(a\) is a positive constant.

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Show that the equation of the tangent to the curve at the point with parameter \(t\) is $$x \sin ^ { 2 } t + y \cos ^ { 2 } t = a \sin ^ { 2 } t \cos ^ { 2 } t$$
3. Hence show that if the tangent meets the \(x\)-axis at \(P\) and the \(y\)-axis at \(Q\), then $$O P + O Q = a$$ where \(O\) is the origin.

6 P is a general point on the curve with parametric equations \(x = 2 t , y = \frac { 2 } { t }\). This is shown in Fig. 6. The tangent at P intersects the \(x\) - and \(y\)-axes at the points Q and R respectively. \begin{figure}[h] \end{figure} Show that the area of the triangle OQR , where O is the origin, is independent of \(t\).

5 A curve is given by the parametric equations \(x = a t ^ { 2 } , y = 2 a\) (where \(a\) is a constant). A point P on the curve has coordinates ( \(a p ^ { 2 }\), 2ap).

1. Find the coordinates of the point, T , where the tangent to the curve at P meets the \(x\)-axis and the coordinates of the point N where the normal to the curve at P meets the \(x\)-axis.
2. Hence show that the area of the triangle PTN is \(2 a ^ { 2 } p \left( p ^ { 2 } + 1 \right)\) square units.

1 Fig. 8 shows a cross-section of a car headlight whose inside reflective surface is modelled, in suitable units, by the curve $$x = 2 t ^ { 2 } , y = 4 t , \quad - \sqrt { 2 } \leqslant t \leqslant \sqrt { 2 } .$$ \(\mathrm { P } \left( 2 t ^ { 2 } , 4 t \right)\) is a point on the curve with parameter \(t\). TS is the tangent to the curve at P , and PR is the line through P parallel to the \(x\)-axis. Q is the point \(( 2,0 )\). The angles that PS and QP make with the positive \(x\)-direction are \(\theta\) and \(\phi\) respectively. \begin{figure}[h] \end{figure}

1. By considering the gradient of the tangent TS, show that \(\tan \theta = \frac { 1 } { t }\).
2. Find the gradient of the line QP in terms of \(t\). Hence show that \(\phi = 2 \theta\), and that angle TPQ is equal to \(\theta\).
   [0pt] [The above result shows that if a lamp bulb is placed at Q , then the light from the bulb is reflected to produce a parallel beam of light.] The inside surface of the headlight has the shape produced by rotating the curve about the \(x\)-axis.
3. Show that the curve has cartesian equation \(y ^ { 2 } = 8 x\). Hence find the volume of revolution of the curve, giving your answer as a multiple of \(\pi\).

7. \begin{figure}[h] \end{figure} A circular tower of radius 1 metre stands in a large horizontal field of grass.A goat is attached to one end of a rope and the other end of the rope is attached to a fixed point \(O\) at the base of the tower.The goat cannot enter the tower. Taking the point \(O\) as the origin( 0,0 ),the centre of the base of the tower is at the point \(T ( 0,1 )\) ,where the unit of length is the metre. The rope has length \(\pi\) metres and you may ignore the size of the goat.
The curve \(C\) shown in Figure 4 represents the edge of the region that the goat can reach.

1. Write down the equation of \(C\) for \(y < 0\) When the goat is at the point \(G ( x , y )\) ,with \(x > 0\) and \(y > 0\) ,as shown in Figure 4 ,the rope lies along \(O A G\) where \(O A\) is an arc of the circle with angle \(O T A = \theta\) radians and \(A G\) is a tangent to the circle at \(A\) .
2. With the aid of a suitable diagram show that $$\begin{aligned} & x = \sin \theta + ( \pi - \theta ) \cos \theta \\ & y = 1 - \cos \theta + ( \pi - \theta ) \sin \theta \end{aligned}$$
3. By considering \(\int y \frac { \mathrm {~d} x } { \mathrm {~d} \theta } \mathrm {~d} \theta\), show that the area, in the first quadrant, between \(C\), the positive \(x\)-axis and the positive \(y\)-axis can be expressed in the form $$\int _ { 0 } ^ { \pi } u \sin u \mathrm {~d} u + \int _ { 0 } ^ { \pi } u ^ { 2 } \sin ^ { 2 } u \mathrm {~d} u + \int _ { 0 } ^ { \pi } u \sin u \cos u \mathrm {~d} u$$
4. Show that \(\int _ { 0 } ^ { \pi } u ^ { 2 } \sin ^ { 2 } u \mathrm {~d} u = \frac { \pi ^ { 3 } } { 6 } + \int _ { 0 } ^ { \pi } u \sin u \cos u \mathrm {~d} u\)
5. Hence find the area of grass that can be reached by the goat.

13 Substitute appropriate values of \(t _ { 1 }\) and \(t _ { 2 }\) to verify that \(t _ { 1 } t _ { 2 }\) gives the correct value for the \(y\)-coordinate of the point of intersection of the tangents at the points A and B in Fig. \(\mathbf { C 1 . }\)

7. \begin{figure}[h] \end{figure} Figure 4 shows a shape \(S ( \theta )\) made up of five line segments \(A B , B C , C D , D E\) and \(E A\) . The lengths of the sides are \(A B = B C = 5 \mathrm {~cm} , C D = E A = 3 \mathrm {~cm}\) and \(D E = 7 \mathrm {~cm}\) . Angle \(B A E =\) angle \(B C D = \theta\) radians. The length of each line segment always remains the same but the value of \(\theta\) can be varied so that different symmetrical shapes can be formed,with the added restriction that none of the line segments cross.

1. Sketch \(S ( \pi )\) ,labelling the vertices clearly. The shape \(S ( \phi )\) is a trapezium.
2. Sketch \(S ( \phi )\) and calculate the value of \(\phi\) . The smallest possible value for \(\theta\) is \(\alpha\) ,where \(\alpha > 0\) ,and the largest possible value for \(\theta\) is \(\beta\) , where \(\beta > \pi\) .
3. Show that \(\alpha = \arccos \left( \frac { 29 } { 40 } \right) \cdot \left[ \arccos ( x ) \right.\) is an alternative notation for \(\left. \cos ^ { - 1 } ( x ) \right]\)
4. Find the value of \(\beta\) . The area,in \(\mathrm { cm } ^ { 2 }\) ,of shape \(S ( \theta )\) is \(R ( \theta )\) .
5. Show that for \(\alpha \leqslant \theta < \pi\) $$R ( \theta ) = 15 \sin \theta + \frac { 7 } { 4 } \sqrt { 87 - 120 \cos \theta }$$ Given that this formula for \(R ( \theta )\) holds for \(\alpha \leqslant \theta \leqslant \beta\)
6. show that \(R ( \theta )\) has only one stationary point and that this occurs when \(\theta = \frac { 2 \pi } { 3 }\)
7. find the maximum and minimum values of \(R ( \theta )\). FOR STYLE, CLARITY AND PRESENTATION: 7 MARKS TOTAL FOR PAPER: 100 MARKS
   END

7 A curve, \(C\), is given parametrically by \(x = 2 \cos \theta , y = 3 \sin \theta , 0 < \theta < \frac { 1 } { 2 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 3 } { 2 } \cot \theta\). A tangent to \(C\) intersects the \(x\)-axis and \(y\)-axis at \(P\) and \(Q\) respectively.
2. Show that the midpoint of \(P Q\) has coordinates \(\left( \sec \theta , \frac { 3 } { 2 } \operatorname { cosec } \theta \right)\).
3. Hence show that the midpoint of \(P Q\) lies on the curve \(\frac { 4 } { x ^ { 2 } } + \frac { 9 } { y ^ { 2 } } = 4\).

A curve is given parametrically by the equations $$x = 4\cos t, \quad y = 3\sin t,$$ where \(0 \leq t \leq \frac{1}{2}\pi\).

1. Find \(\frac{dy}{dx}\) in terms of \(t\). [3]
2. Show that the equation of the tangent at the point \(P\), where \(t = p\), is $$3x\cos p + 4y\sin p = 12.$$ [3]
3. The tangent at \(P\) meets the \(x\)-axis at \(R\) and the \(y\)-axis at \(S\). \(O\) is the origin. Show that the area of triangle \(ORS\) is \(\frac{6}{\sin 2p}\). [3]
4. Write down the least possible value of the area of triangle \(ORS\), and give the corresponding value of \(p\). [3]

Fig. 8 shows a cross-section of a car headlight whose inside reflective surface is modelled, in suitable units, by the curve $$x = 2t^2, y = 4t, \quad -\sqrt{2} < t < \sqrt{2}.$$ P\((2t^2, 4t)\) is a point on the curve with parameter \(t\). TS is the tangent to the curve at P, and PR is the line through P parallel to the \(x\)-axis. Q is the point (2, 0). The angles that PS and QP make with the positive \(x\)-direction are \(\theta\) and \(\phi\) respectively. \includegraphics{figure_8}

1. By considering the gradient of the tangent TS, show that \(\tan \theta = \frac{1}{t}\). [3]
2. Find the gradient of the line QP in terms of \(t\). Hence show that \(\phi = 2\theta\), and that angle TPQ is equal to \(\theta\). [8]

[The above result shows that if a lamp bulb is placed at Q, then the light from the bulb is reflected to produce a parallel beam of light.] The inside surface of the headlight has the shape produced by rotating the curve about the \(x\)-axis.

1. Show that the curve has cartesian equation \(y^2 = 8x\). Hence find the volume of revolution of the curve, giving your answer as a multiple of \(\pi\). [7]

\includegraphics{figure_1} Figure 1 shows the curve with parametric equations $$x = a\sqrt{t}, \quad y = at(1-t), \quad t \geq 0,$$ where \(a\) is a positive constant.

1. Find \(\frac{dy}{dx}\) in terms of \(t\). [3]

The curve meets the \(x\)-axis at the origin, \(O\), and at the point \(A\). The tangent to the curve at \(A\) meets the \(y\)-axis at the point \(B\) as shown.

1. Show that the area of triangle \(OAB\) is \(a^2\). [6]

\includegraphics{figure_5} The diagram shows the curve with parametric equations $$x = a\sqrt{t}, \quad y = at(1-t), \quad t \geq 0,$$ where \(a\) is a positive constant.

1. Find \(\frac{dy}{dx}\) in terms of \(t\). [3]

The curve meets the \(x\)-axis at the origin, \(O\), and at the point \(A\). The tangent to the curve at \(A\) meets the \(y\)-axis at the point \(B\) as shown.

1. Show that the area of triangle \(OAB\) is \(a^2\). [5]

The curve defined by the parametric equations $$x = t^2 \text{ and } y = 2t \quad -\sqrt{2} \leq t \leq \sqrt{2}$$ is shown in Figure 1 below. \includegraphics{figure_1}

1. Find a Cartesian equation of the curve in the form \(y^2 = f(x)\) [2 marks]
2. 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 \(AB\) is at an angle \(\phi\) to the \(x\)-axis. \includegraphics{figure_1_extended}
   1. By considering the gradient of the curve, show that $$\tan \theta = \frac{1}{a}$$ [3 marks]
   2. Find \(\tan \phi\) in terms of \(a\). [2 marks]
   3. Show that \(\tan 2\theta = \tan \phi\) [3 marks]

A curve is defined, for \(t \geqslant 0\), by the parametric equations $$x = t^2, \quad y = t^3.$$

1. Show that the equation of the tangent at the point with parameter \(t\) is $$2y = 3tx - t^3.$$ [4]

1. In this question you must show detailed reasoning. It is given that this tangent passes through the point \(A\left(\frac{19}{2}, -\frac{15}{8}\right)\) and it meets the \(x\)-axis at the point \(B\). Find the area of triangle \(OAB\), where \(O\) is the origin. [7]

The curve \(C\) has parametric equations $$x = \cos^2 t$$ $$y = \cos t \sin t$$ where \(0 \leq t < \pi\)

1. Show that \(C\) is a circle and find its centre and its radius. [5]

% Figure 1 shows a sketch of C with point P, rectangle R with diagonal OP \includegraphics{figure_1} Figure 1 Figure 1 shows a sketch of \(C\). The point \(P\), with coordinates \((\cos^2 \alpha, \cos\alpha \sin \alpha)\), \(0 < \alpha < \frac{\pi}{2}\), lies on \(C\). The rectangle \(R\) has one side on the \(x\)-axis, one side on the \(y\)-axis and \(OP\) as a diagonal, where \(O\) is the origin.

1. Show that the area of \(R\) is \(\sin\alpha \cos^3 \alpha\) [1]
2. Find the maximum area of \(R\), as \(\alpha\) varies. [7]

[Total 13 marks]