A2 June 2020 Paper 1 Q5
5 \(H_1\) is the locus of points such that the distance from the point \((5, 0)\) is twice the distance from the line \(x = 2\)
(a) Show that the equation of \(H_1\) can be written in the form\[(x - 1)^2 - \frac{y^2}{q} = r\]
where \(q\) and \(r\) are integers. [5 marks]
(b) \(H_2\) is the hyperbola\[x^2 - y^2 = 4\]
Describe fully a sequence of two transformations which maps the graph of \(H_2\) onto the graph of \(H_1\) [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Forms an expression for the distance of a general point from the line \(x = 2\). Condone \((x - 2)\) or \((2 - x)\) | B1 | 1.1b |
| Forms an expression for the (squared) distance of a general point from \((5, 0)\) | B1 | 1.1b |
| Forms an equation in \(x\) and \(y\) for the locus, of the form: distance from \((5, 0)\) = 2 x distance from \(x = 2\) | M1 | 3.1a |
| Simplifies their equation to a quadratic in \(x\) and \(y\) with coefficient of \(x\) equal to 1 | M1 | 1.1a |
| Completes a rigorous argument to show \((x - 1)^2 - \frac{y^2}{3} = 4\) AG | R1 | 2.1 |
Typical solution
\[\begin{aligned} \sqrt{(x - 5)^2 + y^2} &= 2|x - 2| \\ (x - 5)^2 + y^2 &= 4(x - 2)^2 \\ x^2 - 10x + 25 + y^2 &= 4x^2 - 16x + 16 \\ 3x^2 - 6x - y^2 &= 9 \\ x^2 - 2x - \frac{y^2}{3} &= 3 \\ (x - 1)^2 - 1 - \frac{y^2}{3} &= 3 \\ (x - 1)^2 - \frac{y^2}{3} &= 4 \end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| States that there is a horizontal (PI by vector with no \(y\)-component) translation. | B1 | 2.2a |
| Describes translation as \(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\) OE | B1 | 2.5 |
| States that there is a stretch parallel to the \(y\)-axis. | B1 | 2.2a |
| States that the scale factor is \(\sqrt{3}\) FT their \(q\) (or \(\sqrt{q}\) if no value of \(q\) given.) | B1F | 2.5 |
| (9 marks) |
Typical solution
Translation by \(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\)
Stretch parallel to \(y\)-axis.
Scale factor \(\sqrt{3}\)