| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Rectangular hyperbola tangent intersection |
| Difficulty | Challenging +1.2 This is a multi-part parabola question requiring knowledge of focus-directrix properties, tangent equations, and coordinate geometry. Part (a) is straightforward application of the focus formula. Part (b) requires finding tangent equations using parametric forms and solving simultaneously—standard FP1 technique. Part (c) involves algebraic manipulation to verify a geometric relationship. While it requires several steps and careful algebra, all techniques are standard FP1 content with no novel insights needed. Slightly above average difficulty due to length and algebraic complexity, but well within expected FP1 scope. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(SP=\sqrt{(3p^2-a)^2+36p^2}\), with \(a=3\) | M1, B1 | M1: Uses distance formula. B1: States/uses focus at \((3,0)\) or \(a=3\) or directrix \(x=-3\) |
| \(SP=\sqrt{9p^4+18p^2+9} = 3(1+p^2)\) given answer | A1* | cso |
| ALT: Perpendicular distance from \(P\) to directrix \(= SP\); directrix \(x=-3\); so \(SP=3+3p^2=3(1+p^2)\) | M1, B1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(y^2=12x \Rightarrow 2y\frac{dy}{dx}=12\) or \(y=\sqrt{12x}\Rightarrow\frac{dy}{dx}=\sqrt{3}x^{-\frac{1}{2}}\) | M1 | Calculus method for gradient, substitutes \(x\) value at either point |
| Tangent at \(P\) has gradient \(=\frac{1}{p}\) | A1 | Either correct, accept unsimplified |
| Equation: \(y-6p=\frac{1}{p}(x-3p^2)\) or \(py=x+3p^2\) | A1 | One equation of tangent correct |
| Tangent at \(Q\): \(y-6q=\frac{1}{q}(x-3q^2)\) or \(qy=x+3q^2\) | B1 | Both tangent equations correct |
| Eliminate \(x\) or \(y\): \(x=3pq\) or \(y=3(p+q)=3p+3q\) | M1 A1 | M1: Eliminate. A1: First variable |
| \(x=3pq\) and \(y=3(p+q)=3p+3q\) | M1 A1 | M1: Substitute/eliminate again. A1: Both variables correct |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(SR^2=(3-3pq)^2+(3p+3q)^2\ (=9+9p^2q^2+9p^2+9q^2)\) | M1 | Find \(SR^2\) |
| \(SP\cdot SQ=3(1+p^2)\cdot3(1+q^2)\ (=9+9p^2q^2+9p^2+9q^2)\) | M1 | Find \(SP\cdot SQ\) |
| So \(SR^2=SP\cdot SQ\) as required | A1 | Deduce equal after no errors; concluding statement required cso |
# Question 8:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $SP=\sqrt{(3p^2-a)^2+36p^2}$, with $a=3$ | M1, B1 | M1: Uses distance formula. B1: States/uses focus at $(3,0)$ or $a=3$ or directrix $x=-3$ |
| $SP=\sqrt{9p^4+18p^2+9} = 3(1+p^2)$ **given answer** | A1* | cso |
| **ALT:** Perpendicular distance from $P$ to directrix $= SP$; directrix $x=-3$; so $SP=3+3p^2=3(1+p^2)$ | M1, B1, A1 | |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $y^2=12x \Rightarrow 2y\frac{dy}{dx}=12$ or $y=\sqrt{12x}\Rightarrow\frac{dy}{dx}=\sqrt{3}x^{-\frac{1}{2}}$ | M1 | Calculus method for gradient, substitutes $x$ value at either point |
| Tangent at $P$ has gradient $=\frac{1}{p}$ | A1 | Either correct, accept unsimplified |
| Equation: $y-6p=\frac{1}{p}(x-3p^2)$ or $py=x+3p^2$ | A1 | One equation of tangent correct |
| Tangent at $Q$: $y-6q=\frac{1}{q}(x-3q^2)$ or $qy=x+3q^2$ | B1 | Both tangent equations correct |
| Eliminate $x$ or $y$: $x=3pq$ or $y=3(p+q)=3p+3q$ | M1 A1 | M1: Eliminate. A1: First variable |
| $x=3pq$ and $y=3(p+q)=3p+3q$ | M1 A1 | M1: Substitute/eliminate again. A1: Both variables correct |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $SR^2=(3-3pq)^2+(3p+3q)^2\ (=9+9p^2q^2+9p^2+9q^2)$ | M1 | Find $SR^2$ |
| $SP\cdot SQ=3(1+p^2)\cdot3(1+q^2)\ (=9+9p^2q^2+9p^2+9q^2)$ | M1 | Find $SP\cdot SQ$ |
| So $SR^2=SP\cdot SQ$ as required | A1 | Deduce equal after no errors; concluding statement required cso |\begin{enumerate}
\item The point $P \left( 3 p ^ { 2 } , 6 p \right)$ lies on the parabola with equation $y ^ { 2 } = 12 x$ and the point $S$ is the focus of this parabola.\\
(a) Prove that $S P = 3 \left( 1 + p ^ { 2 } \right)$
\end{enumerate}
The point $Q \left( 3 q ^ { 2 } , 6 q \right) , p \neq q$, also lies on this parabola.\\
The tangent to the parabola at the point $P$ and the tangent to the parabola at the point $Q$ meet at the point $R$.\\
(b) Find the equations of these two tangents and hence find the coordinates of the point $R$, giving the coordinates in their simplest form.\\
(c) Prove that $S R ^ { 2 } = S P \cdot S Q$
\hfill \mbox{\textit{Edexcel FP1 2015 Q8 [14]}}