FP1 January 2009 Q8
8. A parabola has equation \(y^2 = 4ax,\ a > 0\). The point \(Q\,(aq^2,\ 2aq)\) lies on the parabola.
(a) Show that an equation of the tangent to the parabola at \(Q\) is \[yq = x + aq^2.\] (4)
This tangent meets the \(y\)-axis at the point \(R\).
(b) Find an equation of the line \(l\) which passes through \(R\) and is perpendicular to the tangent at \(Q\). (3)
(c) Show that \(l\) passes through the focus of the parabola. (1)
(d) Find the coordinates of the point where \(l\) meets the directrix of the parabola. (2)
| Scheme | Marks |
|---|---|
| \(\dfrac{dy}{dx} = a^{\frac{1}{2}}x^{-\frac{1}{2}}\) or \(2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4a\) | M1 |
| The gradient of the tangent is \(\dfrac{1}{q}\) | A1 |
| The equation of the tangent is \(y - 2aq = \dfrac{1}{q}(x - aq^2)\) | M1 |
| So \(yq = x + aq^2\) * | A1 |
| (4) |
Notes
(a) \(\dfrac{dy}{dx} = \dfrac{2a}{2aq}\) OK for M1
Use of \(y = mx + c\) to find \(c\) OK for second M1
Correct solution only for final A1
| Scheme | Marks |
|---|---|
| \(R\) has coordinates \((0,\ aq)\) | B1 |
| The line \(l\) has equation \(y - aq = -qx\) | M1A1 |
| (3) |
Notes
(b) −1/(their gradient in part a) in equation OK for M1
| Scheme | Marks |
|---|---|
| When \(y = 0\), \(x = a\) (so line \(l\) passes through \((a,\ 0)\) the focus of the parabola.) | B1 |
| (1) |
Notes
(c) They must attempt \(y = 0\) or \(x = a\) to show correct coordinates of \(R\) for B1
| Scheme | Marks |
|---|---|
| Line \(l\) meets the directrix when \(x = -a\): Then \(y = 2aq\). So coordinates are \((-a,\ 2aq)\) | M1:A1 |
| (2) | |
| [10] |
Notes
(d) Substitute \(x = -a\) for M1.
Both coordinates correct for A1.