FP1 June 2010 Q5
5. The parabola \(C\) has equation \(y^2 = 20x\).
(a) Verify that the point \(P(5t^2,\ 10t)\) is a general point on \(C\). (1)
The point \(A\) on \(C\) has parameter \(t = 4\).
The line \(l\) passes through \(A\) and also passes through the focus of \(C\).
(b) Find the gradient of \(l\). (4)
| Scheme | Marks |
|---|---|
| \(y^2 = (10t)^2 = 100t^2\) and \(20x = 20 \times 5t^2 = 100t^2\) | B1 |
| (1) |
Alternative method
| Scheme | Marks |
|---|---|
| Compare with \(y^2 = 4ax\) and identify \(a = 5\) to give answer. | B1 |
| (1) |
Notes
(a) Allow substitution of \(x\) to obtain \(y = \pm 10t\) (or just \(10t\)) or of \(y\) to obtain \(x\)
| Scheme | Marks |
|---|---|
| Point \(A\) is (80, 40) (stated or seen on diagram). May be given in part (a) | B1 |
| Focus is (5, 0) (stated or seen on diagram) or \((a,\ 0)\) with \(a = 5\) May be given in part (a). | B1 |
| Gradient: \(\dfrac{40 - 0}{80 - 5} = \dfrac{40}{75} \ \left(= \dfrac{8}{15}\right)\) | M1 A1 |
| (4) | |
| 5 marks |
Notes
(b) M1: requires use of gradient formula correctly, for their values of \(x\) and \(y\).
This mark may be implied by correct answer.
Differentiation is M0 A0
A1: Accept 0.533 or awrt