A2 June 2022 Q5
5. The rectangular hyperbola \(H\) has equation \(xy = 36\)
The point \(Q\left(12t, \dfrac{3}{t}\right)\) also lies on \(H\).
The tangent at \(P\) and the tangent at \(Q\) meet at the point \(R\).
| Scheme | Marks | AO |
|---|---|---|
| \(y + x\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{y}{x} = \dfrac{-\frac{6}{t}}{6t} = -\dfrac{1}{t^2}\) or \(y = \dfrac{36}{x} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{36}{x^2} = -\dfrac{36}{(6t)^2} = -\dfrac{1}{t^2}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t} \div \dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{-6t^{-2}}{6} = -\dfrac{1}{t^2}\) | M1 | 1.1b |
| \(y - \dfrac{6}{t} = \text{“}-\dfrac{1}{t^2}\text{”}(x - 6t)\) | M1 | 1.1b |
| \(yt^2 + x = 12t\ *\) | A1* | 2.1 |
| (3) |
Notes
M1: Differentiates implicitly, directly or parametrically to find the gradient at the point \(P\) in terms of \(t\). Allow slips in coefficients, as long as method is clear.
M1: Finds the equation of the tangent at the point \(P\) using their gradient (not reciprocal etc). If using \(y = mx + c\) must proceed to find \(c\) and substitute back in to equation.
A1*: The correct equation for the tangent at the point \(P\) from correct working.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{y}{x} = \dfrac{-\frac{3}{t}}{12t} = -\dfrac{1}{4t^2}\) and \(y - \dfrac{3}{t} = \text{‘}-\dfrac{1}{4t^2}\text{’}(x - 12t)\) | M1 | 1.1b |
| \(y - \dfrac{3}{t} = -\dfrac{1}{4t^2}(x - 12t)\) o.e such as \(4yt^2 + x = 24t\) | A1 | 1.1b |
| (2) |
Notes
M1: Finds the new gradient (any method as above) and proceeds to find the equation of the tangent at the point \(Q\). Alternatively replaces \(t\) by \(2t\) in the answer to (a).
A1: Correct equation - any form, need not be simplified and isw after a correct equation.
| Scheme | Marks | AO |
|---|---|---|
| E.g. \(\left.\begin{aligned}4yt^2 + x &= 24t\\ yt^2 + x &= 12t\end{aligned}\right\}\ 3yt^2 = 12t \Rightarrow y = \ldots\) and \(x = 12t - yt^2 = \ldots\) | M1 | 2.1 |
| \(x = 8t\) and \(y = \dfrac{4}{t}\) | A1 | 1.1b |
| \(xy = \ldots\) | dM1 | 1.1b |
| \(xy = 32\) hence rectangular hyperbola | A1 | 2.4 |
| (4) | ||
| (9 marks) |
Notes
M1: Solves their simultaneous equations to find both the \(x\) and \(y\) coordinate for the point \(R\).
A1: Correct point of intersection, it does not need to be simplified.
dM1: Dependent on the first method mark. Multiplies \(x\) by \(y\) to reach a constant.
A1: Shows that \(xy = 32\) and hence rectangular hyperbola