Find stationary points

5 Given that \(y ^ { 3 } = x y - x ^ { 2 }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - 2 x } { 3 y ^ { 2 } - x }\).
Hence show that the curve \(y ^ { 3 } = x y - x ^ { 2 }\) has a stationary point when \(x = \frac { 1 } { 8 }\).

2 Fig. 9 shows the curve with equation \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\). It has an asymptote \(x = a\) and turning point P . \begin{figure}[h] \end{figure}

1. Write down the value of \(a\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x ^ { 3 } - 3 x ^ { 2 } } { 3 y ^ { 2 } ( 2 x - 1 ) ^ { 2 } }\). Hence find the coordinates of the turning point P , giving the \(y\)-coordinate to 3 significant figures.
3. Show that the substitution \(u = 2 x - 1\) transforms \(\int \frac { x } { \sqrt [ 3 ] { 2 x - 1 } } \mathrm {~d} x\) to \(\frac { 1 } { 4 } \int \left( u ^ { \frac { 2 } { 3 } } + u ^ { - \frac { 1 } { 3 } } \right) \mathrm { d } u\). Hence find the exact area of the region enclosed by the curve \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4.5\).

2 A curve has equation \(x ^ { 2 } + 2 y ^ { 2 } = 4 x\).

1. By differentiating implicitly, find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
   [0pt]
2. Hence find the exact coordinates of the stationary points of the curve. [You need not determine their nature.]

4 Fig. 7 shows the curve \(x ^ { 3 } + y ^ { 3 } = 3 x y\). The point P is a turning point of the curve. \begin{figure}[h] \end{figure}

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - x ^ { 2 } } { y ^ { 2 } - x }\).
2. Hence find the exact \(x\)-coordinate of P .

4 A curve has equation \(2 y ^ { 2 } + y = 9 x ^ { 2 } + 1\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). Hence find the gradient of the curve at the point \(\mathrm { A } ( 1,2 )\).
2. Find the coordinates of the points on the curve at which \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\).

1 Fig. 9 shows the curve with equation \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\). It has an asymptote \(x = a\) and turning point P . \begin{figure}[h] \end{figure}

1. Write down the value of \(a\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x ^ { 3 } - 3 x ^ { 2 } } { 3 y ^ { 2 } ( 2 x - 1 ) ^ { 2 } }\). Hence find the coordinates of the turning point P , giving the \(y\)-coordinate to 3 significant figures.
3. Show that the substitution \(u = 2 x - 1\) transforms \(\int \frac { x } { \sqrt [ 3 ] { 2 x - 1 } } \mathrm {~d} x\) to \(\frac { 1 } { 4 } \int \left( u ^ { \frac { 2 } { 3 } } + u ^ { - \frac { 1 } { 3 } } \right) \mathrm { d } u\). Hence find the exact area of the region enclosed by the curve \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4.5\).

3 The equation of a curve is \(x ^ { 2 } y - x y ^ { 2 } = 2\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y ^ { 2 } - 2 x y } { x ^ { 2 } - 2 x y }\).
2. (a) Show that, if \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\), then \(y = 2 x\).
   (b) Hence find the coordinates of the point on the curve where the tangent is parallel to the \(x\)-axis.

5

1. For the curve \(2 x ^ { 2 } + x y + y ^ { 2 } = 14\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
2. Deduce that there are two points on the curve \(2 x ^ { 2 } + x y + y ^ { 2 } = 14\) at which the tangents are parallel to the \(x\)-axis, and find their coordinates.

2. A curve has the equation $$2 x ^ { 2 } + x y - y ^ { 2 } + 18 = 0$$ Find the coordinates of the points where the tangent to the curve is parallel to the \(x\)-axis.

9 Fig. 9 shows the curve with equation \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\). It has an asymptote \(x = a\) and turning point P . \begin{figure}[h] \end{figure}

1. Write down the value of \(a\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x ^ { 3 } - 3 x ^ { 2 } } { 3 y ^ { 2 } ( 2 x - 1 ) ^ { 2 } }\). Hence find the coordinates of the turning point P , giving the \(y\)-coordinate to 3 significant figures.
3. Show that the substitution \(u = 2 x - 1\) transforms \(\int \frac { x } { \sqrt [ 3 ] { 2 x - 1 } } \mathrm {~d} x\) to \(\frac { 1 } { 4 } \int \left( u ^ { \frac { 2 } { 3 } } + u ^ { - \frac { 1 } { 3 } } \right) \mathrm { d } u\). Hence find the exact area of the region enclosed by the curve \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4.5\).

8 The equation of a curve is \(x ^ { 3 } + y ^ { 3 } = 6 x y\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
2. Show that the point \(\left( 2 ^ { \frac { 4 } { 3 } } , 2 ^ { \frac { 5 } { 3 } } \right)\) lies on the curve and that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) at this point.
3. The point \(( a , a )\), where \(a > 0\), lies on the curve. Find the value of \(a\) and the gradient of the curve at this point.

3 The equation of a curve is \(x y ^ { 2 } = x ^ { 2 } + 1\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\), and hence find the coordinates of the stationary points on the curve.

5 Find the coordinates of the two stationary points on the curve with equation $$x ^ { 2 } + 4 x y + 2 y ^ { 2 } + 18 = 0$$

7 The curve \(C\) has equation \(x ^ { 2 } + 2 x y - 4 y ^ { 2 } + 20 = 0\). Show that if the tangent to \(C\) at the point \(( x , y )\) is parallel to the \(x\)-axis then \(x + y = 0\). Hence find the coordinates of the stationary points on \(C\), and determine their nature.

10 The curve \(C\) has equation \(x ^ { 3 } + y ^ { 3 } = 3 x y\), for \(x > 0\) and \(y > 0\). Find a relationship between \(x\) and \(y\) when \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\). Find the exact coordinates of the turning point of \(C\), and determine the nature of this turning point.

1. A curve \(C\) is given by the equation

$$\sin x + \cos y = 0.5 \quad - \frac { \pi } { 2 } \leqslant x < \frac { 3 \pi } { 2 } , - \pi < y < \pi$$ A point \(P\) lies on \(C\).
The tangent to \(C\) at the point \(P\) is parallel to the \(x\)-axis.
Find the exact coordinates of all possible points \(P\), justifying your answer.
(Solutions based entirely on graphical or numerical methods are not acceptable.)

13 Determine the coordinates of the turning points on the curve with equation $$y ^ { 2 } + x y + x ^ { 2 } - x = 1 .$$

5 A curve is defined implicitly by the equation \(2 x ^ { 2 } + 3 x y + y ^ { 2 } + 2 = 0\).

1. Show that \(\frac { d y } { d x } = - \frac { 4 x + 3 y } { 3 x + 2 y }\).
2. In this question you must show detailed reasoning. Find the coordinates of the stationary points of the curve.

6 The surface \(C\) is given by the equation \(z = x ^ { 2 } + y ^ { 3 } + a x y\) for all real \(x\) and \(y\), where \(a\) is a non-zero real number.

1. Show that \(C\) has two stationary points, one of which is at the origin, and give the coordinates of the second in terms of \(a\).
2. Determine the nature of these stationary points of \(C\).
3. Explain what can be said about the location and nature of the stationary point(s) of the surface given by the equation \(z = x ^ { 2 } + y ^ { 3 }\) for all real \(x\) and \(y\).

6 A curve is defined by the equation \(9 x ^ { 2 } - 6 x y + 4 y ^ { 2 } = 3\). Find the coordinates of the two stationary points of this curve.

5 A curve is defined by the equation $$y ^ { 2 } - x y + 3 x ^ { 2 } - 5 = 0$$

1. Find the \(y\)-coordinates of the two points on the curve where \(x = 1\).
2. 1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - 6 x } { 2 y - x }\).
   2. Find the gradient of the curve at each of the points where \(x = 1\).
   3. Show that, at the two stationary points on the curve, \(33 x ^ { 2 } - 5 = 0\).

13 In this question you must show detailed reasoning. Find the exact values of the \(x\)-coordinates of the stationary points of the curve \(x ^ { 3 } + y ^ { 3 } = 3 x y + 35\).

15 The curve with equation $$x ^ { 2 } + 2 y ^ { 3 } - 4 x y = 0$$ has a single stationary point at \(P\) as shown in the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-26_656_1138_548_450} 15

1. Show that the \(y\)-coordinate of \(P\) satisfies the equation $$y ^ { 2 } ( y - 2 ) = 0$$ 15
2. Hence, find the coordinates of \(P\) [0pt] [2 marks]

9 Find the equations of all the horizontal tangents to the curve with equation \(y ^ { 2 } = x ^ { 4 } - 4 x ^ { 3 } + 36\).

7 Find the coordinates of the two stationary points of the curve $$9 x ^ { 2 } + 4 y ^ { 2 } - 6 x - 4 y = 34$$ showing that one is a maximum and one is a minimum.