| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (polynomial/rational) |
| Difficulty | Standard +0.3 This is a straightforward C4 parametric equations question requiring standard techniques: chain rule differentiation (dy/dx = (dy/dt)/(dx/dt)), finding a normal line, and algebraic manipulation to eliminate the parameter. All three parts follow routine procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks |
|---|---|
| \(\frac{dx}{dt} = \frac{1\times(2-t)-t\times(-1)}{(2-t)^2} = \frac{2}{(2-t)^2}\) | M1 B1 |
| \(\frac{dy}{dt} = -(1+t)^{-2}\) | B1 |
| \(\frac{dy}{dx} = -\frac{1}{(1+t)^2} \div \frac{2}{(2-t)^2} = -\frac{(2-t)^2}{2(1+t)^2} = -\frac{1}{2}\left(\frac{2-t}{1+t}\right)^2\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(t=1,\ x=1,\ y=\frac{1}{2},\ \text{grad} = -\frac{1}{8}\) | B1 |
| grad of normal \(= 8\) | |
| \(y - \frac{1}{2} = 8(x-1)\) giving \(y = 8x - \frac{15}{2}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x(2-t) = t\) | M1 | |
| \(2x = t(1+x),\ t = \frac{2x}{1+x}\) | A1 | |
| \(y = \frac{1}{1+\frac{2x}{1+x}} = \frac{1+x}{(1+x)+2x} \therefore y = \frac{1+x}{1+3x}\) | M1 A1 | (11) |
# Question 6:
## Part (i):
| $\frac{dx}{dt} = \frac{1\times(2-t)-t\times(-1)}{(2-t)^2} = \frac{2}{(2-t)^2}$ | M1 B1 | |
|---|---|---|
| $\frac{dy}{dt} = -(1+t)^{-2}$ | B1 | |
| $\frac{dy}{dx} = -\frac{1}{(1+t)^2} \div \frac{2}{(2-t)^2} = -\frac{(2-t)^2}{2(1+t)^2} = -\frac{1}{2}\left(\frac{2-t}{1+t}\right)^2$ | M1 A1 | |
## Part (ii):
| $t=1,\ x=1,\ y=\frac{1}{2},\ \text{grad} = -\frac{1}{8}$ | B1 | |
|---|---|---|
| grad of normal $= 8$ | | |
| $y - \frac{1}{2} = 8(x-1)$ giving $y = 8x - \frac{15}{2}$ | M1 A1 | |
## Part (iii):
| $x(2-t) = t$ | M1 | |
|---|---|---|
| $2x = t(1+x),\ t = \frac{2x}{1+x}$ | A1 | |
| $y = \frac{1}{1+\frac{2x}{1+x}} = \frac{1+x}{(1+x)+2x} \therefore y = \frac{1+x}{1+3x}$ | M1 A1 | **(11)** |
---6. A curve has parametric equations
$$x = \frac { t } { 2 - t } , \quad y = \frac { 1 } { 1 + t } , \quad - 1 < t < 2$$
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 1 } { 2 } \left( \frac { 2 - t } { 1 + t } \right) ^ { 2 }$.\\
(ii) Find an equation for the normal to the curve at the point where $t = 1$.\\
(iii) Show that the cartesian equation of the curve can be written in the form
$$y = \frac { 1 + x } { 1 + 3 x }$$
\hfill \mbox{\textit{OCR C4 Q6 [11]}}