FP3 June 2013 Q1
1. A hyperbola \(H\) has equation \[\frac{x^2}{a^2} - \frac{y^2}{25} = 1, \qquad \text{where } a \text{ is a positive constant.}\]
The foci of \(H\) are at the points with coordinates \((13, 0)\) and \((-13, 0)\).
Find
(a) the value of the constant \(a\), (3)
(b) the equations of the directrices of \(H\). (3)
| Scheme | Marks |
|---|---|
| Mark (a) and (b) together | |
| \(ae = 13\) and \(a^2(e^2 - 1) = 25\) Sight of both of these (can be implied by their work) (allow \(\pm ae = \pm 13\) or \(\pm ae = 13\) or \(ae = \pm 13\)) | B1 |
| Solves to obtain \(a^2 = \ldots\) or \(a = \ldots\) Eliminates \(e\) to reach \(a^2 = \ldots\) or \(a = \ldots\) | M1 |
| \(a = 12\) Cao (not \(\pm 12\)) unless \(-12\) is rejected | A1 |
| Scheme | Marks |
|---|---|
| \(e = 13/\text{“}12\text{”}\) Uses their \(a\) to find \(e\) or finds \(e\) by eliminating \(a\) (Ignore \(\pm\) here) (Can be implied by a correct answer) | M1 |
| \(x = (\pm)\dfrac{a}{e},\ = \pm\dfrac{144}{13}\) | M1, A1 |
| (6 marks) |
Notes
M1: \((x = )(\pm)\dfrac{a}{e}\) \(\pm\) not needed for this mark nor is \(x\) and even allow \(y = (\pm)\dfrac{a}{e}\) here – just look for use of \(\dfrac{a}{e}\) with numerical \(a\) and \(e\).
A1: \(x = \pm\dfrac{144}{13}\) oe but must be an equation (Do not allow \(x = \pm\dfrac{12}{13/12}\))
If they use the eccentricity equation for the ellipse \((b^2 = a^2(1 - e^2))\) allow the M’s