FP3 June 2009 Q6
6. The hyperbola \(H\) has equation \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\), where \(a\) and \(b\) are constants.
The line \(L\) has equation \(y = mx + c\), where \(m\) and \(c\) are constants.
(a) Given that \(L\) and \(H\) meet, show that the \(x\)-coordinates of the points of intersection are the roots of the equation \[(a^2m^2 - b^2)x^2 + 2a^2mcx + a^2(c^2 + b^2) = 0\] (2)
Hence, given that \(L\) is a tangent to \(H\),
(b) show that \(\quad a^2m^2 = b^2 + c^2\). (2)
The hyperbola \(H'\) has equation \(\dfrac{x^2}{25} - \dfrac{y^2}{16} = 1\).
(c) Find the equations of the tangents to \(H'\) which pass through the point \((1, 4)\). (7)
| Scheme | Marks |
|---|---|
| \(\dfrac{x^2}{a^2} - \dfrac{(mx + c)^2}{b^2} = 1\) and so \(b^2x^2 - a^2(mx + c)^2 = a^2b^2\) | M1 |
| \(\therefore (b^2 - a^2m^2)x^2 - 2a^2mcx - a^2(c^2 + b^2) = 0\) Or \((a^2m^2 - b^2)x^2 + 2a^2mcx + a^2(c^2 + b^2) = 0\) * | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \((2a^2mc)^2 = 4(a^2m^2 - b^2) \times a^2(c^2 + b^2)\) \(4a^4m^2c^2 = -4a^2(b^2c^2 + b^4 - a^2m^2c^2 - a^2m^2b^2)\) | M1 |
| \(c^2 = a^2m^2 - b^2\) or \(a^2m^2 = b^2 + c^2\) * | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| Substitute \((1, 4)\) into \(y = mx + c\) to give \(4 = m + c\) and Substitute \(a = 5\) and \(b = 4\) into \(c^2 = a^2m^2 - b^2\) to give \(c^2 = 25m^2 - 16\) | B1 |
| Solve simultaneous equations to eliminate \(m\) or \(c\): \((4 - m)^2 = 25m^2 - 16\) | M1 |
| To obtain \(24m^2 + 8m - 32 = 0\) | A1 |
| Solve to obtain \(8(3m + 4)(m - 1) = 0 \ldots\ldots m = \ldots\text{or}\ldots\) | M1 |
| \(m = 1\) or \(-\dfrac{4}{3}\) | A1 |
| Substitute to get \(c = 3\) or \(\dfrac{16}{3}\) | M1 |
| Lines are \(y = x + 3\) and \(3y + 4x = 16\) | A1 |
| (7) | |
| (11 marks) |