1.07s Parametric and implicit differentiation

A parabola has equation \(y^2 = 4ax\), \(a > 0\). The point \(Q (aq^2, 2aq)\) lies on the parabola.

1. Show that an equation of the tangent to the parabola at \(Q\) is $$yq = x + aq^2.$$ [4]
2. This tangent meets the \(y\)-axis at the point \(R\). Find an equation of the line \(l\) which passes through \(R\) and is perpendicular to the tangent at \(Q\). [3]
3. Show that \(l\) passes through the focus of the parabola. [1]
4. Find the coordinates of the point where \(l\) meets the directrix of the parabola. [2]

The parabola \(C\) has equation \(y^2 = 4ax\), where \(a\) is a positive constant. The point \(P(at^2, 2at)\) is a general point on \(C\).

1. Show that the equation of the tangent to \(C\) at \(P(at^2, 2at)\) is $$ty = x + at^2$$ [4]

The tangent to \(C\) at \(P\) meets the \(y\)-axis at a point \(Q\).

1. Find the coordinates of \(Q\). [1]

Given that the point \(S\) is the focus of \(C\),

1. show that \(PQ\) is perpendicular to \(SQ\). [3]

The point \(P \left( 2p, \frac{2}{p} \right)\) and the point \(Q \left( 2q, \frac{2}{q} \right)\), where \(p \neq -q\), lie on the rectangular hyperbola with equation \(xy = 4\). The tangents to the curve at the points \(P\) and \(Q\) meet at the point \(R\). Show that at the point \(R\), $$x = \frac{4pq}{p + q} \text{ and } y = \frac{4}{p + q}.$$ [8]

Show that the normal to the rectangular hyperbola \(xy = c^2\), at the point \(P \left( ct, \frac{c}{t} \right)\), \(t \neq 0\) has equation $$y = t^2 x + \frac{c}{t} - ct^3.$$ [5]

The curve \(C\) has polar equation $$r = 1 + 2 \cos \theta, \quad 0 \leq \theta \leq \frac{\pi}{2}.$$ At the point \(P\) on \(C\), the tangent to \(C\) is parallel to the initial line. Given that \(O\) is the pole, find the exact length of the line \(OP\). [7]

$$x \frac{dy}{dx} = 3x + y^2.$$

1. Show that $$\frac{d^2 y}{dx^2} + (1 - 2y) \frac{dy}{dx} = 3.$$ [2]

Given that \(y = 1\) at \(x = 1\),

1. find a series solution for \(y\) in ascending powers of \((x - 1)\), up to and including the term in \((x - 1)^3\). [8]

The hyperbola \(H\) has equation $$4x^2 - y^2 = 4$$

1. Write down the equations of the asymptotes of \(H\). [1]
2. Find the coordinates of the foci of \(H\). [2]

The point \(P(\sec \theta, 2 \tan \theta)\) lies on \(H\).

1. Using calculus, show that the equation of the tangent to \(H\) at the point \(P\) is $$y \tan \theta = 2x \sec \theta - 2$$ [4]

The point \(V(-1, 0)\) and the point \(W(1, 0)\) both lie on \(H\). The point \(Q(\sec \theta, -2 \tan \theta)\) also lies on \(H\). Given that \(P\), \(Q\), \(V\) and \(W\) are distinct points on \(H\) and that the lines \(VP\) and \(WQ\) intersect at the point \(S\),

1. show that, as \(\theta\) varies, \(S\) lies on an ellipse with equation $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ where \(a\) and \(b\) are integers to be found. [7]

An ellipse has equation $$\frac{x^2}{25} + \frac{y^2}{4} = 1$$ The point \(P\) lies on the ellipse and has coordinates \((5\cos \theta, 2\sin \theta)\), \(0 < \theta < \frac{\pi}{2}\) The line \(L\) is a normal to the ellipse at the point \(P\).

1. Show that an equation for \(L\) is $$5x \sin \theta - 2y \cos \theta = 21 \sin \theta \cos \theta$$ [5]

Given that the line \(L\) crosses the \(y\)-axis at the point \(Q\) and that \(M\) is the midpoint of \(PQ\),

1. find the exact area of triangle \(OPM\), where \(O\) is the origin, giving your answer as a multiple of \(\sin 2\theta\) [6]

The hyperbola \(H\) has equation $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

1. Use calculus to show that the equation of the tangent to \(H\) at the point \((a\cosh\theta, b\sinh\theta)\) may be written in the form $$xb\cosh\theta - ya\sinh\theta = ab$$ [4] The line \(l_1\) is the tangent to \(H\) at the point \((a\cosh\theta, b\sinh\theta)\), \(\theta \neq 0\). Given that \(l_1\) meets the \(x\)-axis at the point \(P\),
2. find, in terms of \(a\) and \(\theta\), the coordinates of \(P\). [2] The line \(l_2\) is the tangent to \(H\) at the point \((a, 0)\). Given that \(l_1\) and \(l_2\) meet at the point \(Q\),
3. find, in terms of \(a\), \(b\) and \(\theta\), the coordinates of \(Q\). [2]
4. Show that, as \(\theta\) varies, the locus of the mid-point of \(PQ\) has equation $$x(4y^2 + b^2) = ab^2$$ [6]

A circle \(C\) with centre \(O\) and radius \(r\) has cartesian equation \(x^2 + y^2 = r^2\) where \(r\) is a constant.

1. Show that \(1 + \left(\frac{dy}{dx}\right)^2 = \frac{r^2}{r^2 - x^2}\) [3]
2. Show that the surface area of the sphere generated by rotating \(C\) through \(\pi\) radians about the \(x\)-axis is \(4\pi r^2\). [5]
3. Write down the length of the arc of the curve \(y = \sqrt{1 - x^2}\) from \(x = 0\) to \(x = 1\) [1]

The hyperbola \(C\) has equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).

1. Show that an equation of the normal to \(C\) at the point \(P(a \sec t, b \tan t)\) is $$ax \sin t + by = (a^2 + b^2) \tan t.$$ [6]

The normal to \(C\) at \(P\) cuts the \(x\)-axis at the point \(A\) and \(S\) is a focus of \(C\). Given that the eccentricity of \(C\) is \(\frac{3}{2}\), and that \(OA = 3OS\), where \(O\) is the origin,

1. determine the possible values of \(t\), for \(0 \leq t < 2\pi\). [8]

The hyperbola \(C\) has equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)

1. Show that an equation of the normal to \(C\) at \(P(a \sec \theta, b \tan \theta)\) is $$by + ax \sin \theta = (a^2 + b^2)\tan \theta$$ [6] The normal at \(P\) cuts the coordinate axes at \(A\) and \(B\). The mid-point of \(AB\) is \(M\).
2. Find, in cartesian form, an equation of the locus of \(M\) as \(\theta\) varies. [7]

(Total 13 marks)

\includegraphics{figure_7} The diagram shows the curve with equation $$x = (y + 4)\ln (2y + 3).$$ The curve crosses the \(x\)-axis at \(A\) and the \(y\)-axis at \(B\).

1. Find an expression for \(\frac{dx}{dy}\) in terms of \(y\). [3]
2. Find the gradient of the curve at each of the points \(A\) and \(B\), giving each answer correct to 2 decimal places. [5]

\includegraphics{figure_3} The diagram shows the curve with equation \(x = (37 + 10y - 2y^2)^{\frac{1}{2}}\).

1. Find an expression for \(\frac{dx}{dy}\) in terms of \(y\). [2]
2. Hence find the equation of the tangent to the curve at the point \((7, 3)\), giving your answer in the form \(y = mx + c\). [5]

Fig. 7 shows the curve defined implicitly by the equation $$y^2 + y = x^3 + 2x,$$ together with the line \(x = 2\). \includegraphics{figure_7} Find the coordinates of the points of intersection of the line and the curve. Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\). Hence find the gradient of the curve at each of these two points.

A curve is defined by the equation \((x + y)^2 = 4x\). The point \((1, 1)\) lies on this curve. By differentiating implicitly, show that \(\frac{dy}{dx} = \frac{2}{x + y} - 1\). Hence verify that the curve has a stationary point at \((1, 1)\). [4]

Fig. 7 shows the curve \(x^3 + y^3 = 3xy\). The point P is a turning point of the curve. \includegraphics{figure_7}

1. Show that \(\frac{dy}{dx} = \frac{y - x^2}{y^2 - x}\). [4]
2. Hence find the exact \(x\)-coordinate of P. [4]

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

1. By differentiating implicitly, find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\). [3]
2. Hence find the exact coordinates of the stationary points of the curve. [You need not determine their nature.] [3]

A curve is defined by the equation \(\sin 2x + \cos y = \sqrt{3}\).

1. Verify that the point P \((\frac{\pi}{6}, \frac{\pi}{6})\) lies on the curve. [1]
2. Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\). Hence find the gradient of the curve at the point P. [5]

Fig. 8 shows the curve \(y = f(x)\), where \(f(x) = \frac{x}{\sqrt{2 + x^2}}\). \includegraphics{figure_8}

1. Show algebraically that \(f(x)\) is an odd function. Interpret this result geometrically. [3]
2. Show that \(f'(x) = \frac{2}{(2 + x^2)^{\frac{3}{2}}}\). Hence find the exact gradient of the curve at the origin. [5]
3. Find the exact area of the region bounded by the curve, the \(x\)-axis and the line \(x = 1\). [4]
4. 1. Show that if \(y = \frac{x}{\sqrt{2 + x^2}}\), then \(\frac{1}{y^2} = \frac{2}{x^2} + 1\). [2]
   2. Differentiate \(\frac{1}{y^2} = \frac{2}{x^2} + 1\) implicitly to show that \(\frac{dy}{dx} = \frac{2y^3}{x^3}\). Explain why this expression cannot be used to find the gradient of the curve at the origin. [4]

Fig. 6 shows part of the curve \(\sin 2y = x - 1\). P is the point with coordinates \((1.5, \frac{1}{12}\pi)\) on the curve. \includegraphics{figure_6}

1. Find \(\frac{dy}{dx}\) in terms of \(y\). Hence find the exact gradient of the curve \(\sin 2y = x - 1\) at the point P. [4]

The part of the curve shown is the image of the curve \(y = \arcsin x\) under a sequence of two geometrical transformations.

1. Find \(y\) in terms of \(x\) for the curve \(\sin 2y = x - 1\). Hence describe fully the sequence of transformations. [4]

You are given that \(y^2 = 4x + 7\).

1. Use implicit differentiation to find \(\frac{dy}{dx}\) in terms of \(y\). [2]
2. Make \(x\) the subject of the equation. Find \(\frac{dx}{dy}\) and hence show that in this case \(\frac{dx}{dy} = \frac{1}{\frac{dx}{dy}}\). [3]

Fig. 3 shows the curve defined by the equation \(y = \arcsin(x - 1)\), for \(0 \leqslant x \leqslant 2\). \includegraphics{figure_7}

1. Find \(x\) in terms of \(y\), and show that \(\frac{dx}{dy} = \cos y\). [3]
2. Hence find the exact gradient of the curve at the point where \(x = 1.5\). [4]

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

1. Find \(\frac{dy}{dx}\) in terms of \(t\). [3 marks]
2. Find an equation of the normal to the curve at the point where \(t = 1\). [4 marks]
3. Find a cartesian equation of the curve. [2 marks]

A curve has equation \(x^3 y + \cos(\pi y) = 7\).

1. Find the exact value of the \(x\)-coordinate at the point on the curve where \(y = 1\). [2 marks]
2. Find the gradient of the curve at the point where \(y = 1\). [5 marks]