Tangent/normal meets curve again

11. The curve \(C\) has parametric equations $$x = 10 \cos 2 t , \quad y = 6 \sin t , \quad - \frac { \pi } { 2 } \leqslant t \leqslant \frac { \pi } { 2 }$$ The point \(A\) with coordinates \(( 5,3 )\) lies on \(C\).

1. Find the value of \(t\) at the point \(A\).
2. Show that an equation of the normal to \(C\) at \(A\) is $$3 y = 10 x - 41$$ The normal to \(C\) at \(A\) cuts \(C\) again at the point \(B\).
3. Find the exact coordinates of \(B\).

8. A curve has parametric equations $$x = t ^ { 2 } - t , \quad y = \frac { 4 t } { 1 - t } \quad t \neq 1$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\), giving your answer as a simplified fraction.
2. Find an equation for the tangent to the curve at the point \(P\) where \(t = - 1\), giving your answer in the form \(a x + b y + c = 0\) where \(a , b\) and \(c\) are integers. The tangent to the curve at \(P\) cuts the curve at the point \(Q\).
3. Use algebra to find the coordinates of \(Q\).

9. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve with parametric equations $$x = t ^ { 2 } + 2 t , \quad y = t ^ { 3 } - 9 t , \quad t \in \mathbb { R }$$ The curve cuts the \(x\)-axis at the origin and at the points \(A\) and \(B\) as shown in Figure 3 .

1. Find the coordinates of point \(A\) and show that point \(B\) has coordinates ( 15,0 ).
2. Show that the equation of the tangent to the curve at \(B\) is \(9 x - 4 y - 135 = 0\) The tangent to the curve at \(B\) cuts the curve again at the point \(X\).
3. Find the coordinates of \(X\).
   (Solutions based entirely on graphical or numerical methods are not acceptable.)

7. The curve \(C\) has parametric equations $$x = 2 \cos t , \quad y = \sqrt { 3 } \cos 2 t , \quad 0 \leqslant t \leqslant \pi$$ where \(t\) is a parameter.

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\). The point \(P\) lies on \(C\) where \(t = \frac { 2 \pi } { 3 }\) The line \(l\) is a normal to \(C\) at \(P\).
2. Show that an equation for \(l\) is $$2 x - 2 \sqrt { 3 } y - 1 = 0$$ The line \(l\) intersects the curve \(C\) again at the point \(Q\).
3. Find the exact coordinates of \(Q\). You must show clearly how you obtained your answers. \includegraphics[max width=\textwidth, alt={}, center]{245bbe52-3a14-4494-af17-7711caf79b22-23_106_63_2595_1882}

1. In this question you must show all stages of your working.

\section*{Solutions relying entirely on calculator technology are not acceptable.} The curve \(C\) has parametric equations $$x = \sin t - 3 \cos ^ { 2 } t \quad y = 3 \sin t + 2 \cos t \quad 0 \leqslant t \leqslant 5$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3\) where \(t = \pi\) The point \(P\) lies on \(C\) where \(t = \pi\)
2. Find the equation of the tangent to the curve at \(P\) in the form \(y = m x + c\) where \(m\) and \(c\) are constants to be found. Given that the tangent to the curve at \(P\) cuts \(C\) at the point \(Q\)
3. show that the value of \(t\) at point \(Q\) satisfies the equation $$9 \cos ^ { 2 } t + 2 \cos t - 7 = 0$$
4. Hence find the exact value of the \(y\) coordinate of \(Q\)

4. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with parametric equations $$x = 2 t ^ { 2 } - 6 t , \quad y = t ^ { 3 } - 4 t , \quad t \in \mathbb { R }$$ The curve cuts the \(x\)-axis at the origin and at the points \(A\) and \(B\), as shown in Figure 2.

1. Find the coordinates of \(A\) and show that \(B\) has coordinates (20, 0).
2. Show that the equation of the tangent to the curve at \(B\) is $$7 y + 4 x - 80 = 0$$ The tangent to the curve at \(B\) cuts the curve again at the point \(P\).
3. Find, using algebra, the \(x\) coordinate of \(P\).

8 A curve has parametric equations $$x = \frac { 1 } { t + 1 } , \quad y = t - 1 .$$ The line \(y = 3 x\) intersects the curve at two points.

1. Show that the value of \(t\) at one of these points is - 2 and find the value of \(t\) at the other point.
2. Find the equation of the normal to the curve at the point for which \(t = - 2\).
3. Find the value of \(t\) at the point where this normal meets the curve again.
4. Find a cartesian equation of the curve, giving your answer in the form \(y = \mathrm { f } ( x )\).

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

$$x = 2 \cos t , \quad y = \sqrt { 3 } \cos 2 t , \quad 0 \leqslant t \leqslant \pi$$

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\). The point \(P\) lies on \(C\) where \(t = \frac { 2 \pi } { 3 }\) The line \(l\) is the normal to \(C\) at \(P\).
2. Show that an equation for \(l\) is $$2 x - 2 \sqrt { 3 } y - 1 = 0$$ The line \(l\) intersects the curve \(C\) again at the point \(Q\).
3. Find the exact coordinates of \(Q\). You must show clearly how you obtained your answers.

5 A curve is defined by the parametric equations \(x = \cos 2 t , y = \sin t\).
The point \(P\) on the curve is where \(t = \frac { \pi } { 6 }\).

1. Find the gradient at \(P\).
2. Find the equation of the normal to the curve at \(P\) in the form \(y = m x + c\).
3. The normal at \(P\) intersects the curve again at the point \(Q ( \cos 2 q , \sin q )\). Use the equation of the normal to form a quadratic equation in \(\sin q\) and hence find the \(x\)-coordinate of \(Q\).
   [0pt] [5 marks]

7 A curve is given parametrically by the equations $$x = t ^ { 2 } , \quad y = \frac { 1 } { t }$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\), giving your answer in its simplest form.
2. Show that the equation of the tangent at the point \(P \left( 4 , - \frac { 1 } { 2 } \right)\) is $$x - 16 y = 12$$
3. Find the value of the parameter at the point where the tangent at \(P\) meets the curve again. June 2005

The parametric equations of a curve are \(x = 2t^2\), \(y = 4t\). Two points on the curve are \(P(2p^2, 4p)\) and \(Q(2q^2, 4q)\).

1. Show that the gradient of the normal to the curve at \(P\) is \(-p\). [2]
2. Show that the gradient of the chord joining the points \(P\) and \(Q\) is \(\frac{2}{p + q}\). [2]
3. The chord \(PQ\) is the normal to the curve at \(P\). Show that \(p^2 + pq + 2 = 0\). [2]
4. The normal at the point \(R(8, 8)\) meets the curve again at \(S\). The normal at \(S\) meets the curve again at \(T\). Find the coordinates of \(T\). [4]

A curve is given parametrically by the equations $$x = t^2, \quad y = \frac{1}{t}.$$

1. Find \(\frac{dy}{dx}\) in terms of \(t\), giving your answer in its simplest form. [3]
2. Show that the equation of the tangent at the point \(P\left(4, -\frac{1}{4}\right)\) is \(x - 16y = 12\). [3]
3. Find the value of the parameter at the point where the tangent at \(P\) meets the curve again. [4]

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 \(\frac{\mathrm{d}y}{\mathrm{d}x}\) 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]