| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2010 |
| Session | June |
| Marks | 14 |
| Topic | Conic sections |
| Difficulty | Standard +0.8 This is a substantial parabola problem requiring derivation from first principles (equidistant definition), coordinate geometry, differentiation, and multi-step algebraic manipulation. Part (iii)(b) involves finding specific coordinates and expressing a length in surd form with a constraint. While the techniques are A-level standard (distance formula, differentiation, tangent equations), the extended reasoning across multiple parts and the need to work with general parameters before specializing makes this moderately challenging, though not requiring exceptional insight. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Using distance-squared, the condition for a point \((x, y)\) to be on the curve is: \((x-0)^2 + (y-a)^2 = (y-b)^2\) | M1 A1 | |
| Expand, simplify and make \(y\) the subject, obtaining: \(y = \frac{1}{2(a-b)}x^2 + \frac{1}{2}(a+b)\) | A1 AG | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| As \(a > b\) the curve has a lowest point with \(y\)-coordinate \(\frac{1}{2}(a+b)\) | B1 | |
| This is the midpoint of the interval on the \(y\)-axis joining \(F\) to the point \((0, b)\) on \(D\). The curve passes between these two points without crossing \(D\) | B1 AG | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| (a) The \(y\)-coordinate of \(P\) is \(a + b\) (or unsimplified expression) | B1 | |
| The gradient of the tangent at \(P\) is \(\frac{x}{a-b}\) | B1 | |
| Attempt at forming the equation of the tangent: \(y - (a+b) = \frac{\sqrt{a^2 - b^2}}{a-b}\left(x - \sqrt{a^2 - b^2}\right)\) | M1 | |
| Which simplifies to: \(y = \frac{\sqrt{a^2 - b^2}}{a-b}x\), passing through the origin | A1 AG | [4] |
| (b) The coordinates of \(P\) are \((4\sqrt{5}, 4)\) | B1 | |
| The coordinates of \(Q\) are \((-8\sqrt{5}, -8)\) | B1 | |
| The length of \(PQ\) is \(\sqrt{144 \times 5 + 144}\) | M1 A1 | |
| This simplifies to \(-12\sqrt{6}\) | A1 | [5] |
### (i)
Using distance-squared, the condition for a point $(x, y)$ to be on the curve is: $(x-0)^2 + (y-a)^2 = (y-b)^2$ | M1 A1 |
Expand, simplify and make $y$ the subject, obtaining: $y = \frac{1}{2(a-b)}x^2 + \frac{1}{2}(a+b)$ | A1 AG | [3]
### (ii)
As $a > b$ the curve has a lowest point with $y$-coordinate $\frac{1}{2}(a+b)$ | B1 |
This is the midpoint of the interval on the $y$-axis joining $F$ to the point $(0, b)$ on $D$. The curve passes between these two points without crossing $D$ | B1 AG | [2]
### (iii)
**(a)** The $y$-coordinate of $P$ is $a + b$ (or unsimplified expression) | B1 |
The gradient of the tangent at $P$ is $\frac{x}{a-b}$ | B1 |
Attempt at forming the equation of the tangent: $y - (a+b) = \frac{\sqrt{a^2 - b^2}}{a-b}\left(x - \sqrt{a^2 - b^2}\right)$ | M1 |
Which simplifies to: $y = \frac{\sqrt{a^2 - b^2}}{a-b}x$, passing through the origin | A1 AG | [4]
**(b)** The coordinates of $P$ are $(4\sqrt{5}, 4)$ | B1 |
The coordinates of $Q$ are $(-8\sqrt{5}, -8)$ | B1 |
The length of $PQ$ is $\sqrt{144 \times 5 + 144}$ | M1 A1 |
This simplifies to $-12\sqrt{6}$ | A1 | [5]The point $F$ has coordinates $(0, a)$ and the straight line $D$ has equation $y = b$, where $a$ and $b$ are constants with $a > b$. The curve $C$ consists of points equidistant from $F$ and $D$.
\begin{enumerate}[label=(\roman*)]
\item Show that the cartesian equation of $C$ can be expressed in the form
$$y = \frac{1}{2(a-b)}x^2 + \frac{1}{2}(a+b).$$ [3]
\item State the $y$-coordinate of the lowest point of the curve and prove that $F$ and $D$ are on opposite sides of $C$. [2]
\item \begin{enumerate}[label=(\alph*)]
\item The point $P$ on the curve has $x$-coordinate $\sqrt{a^2 - b^2}$, where $|a| > |b|$. Show that the tangent at $P$ passes through the origin. [4]
\item The tangent at $P$ intersects the line $D$ at the point $Q$. In the case that $a = 12$ and $b = -8$, find the coordinates of $P$ and $Q$. Show that the length of $PQ$ can be expressed as $p\sqrt{q}$, where $p = 2q$. [5]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2010 Q8 [14]}}