FP3 June 2013 Q7
7. The ellipse \(E\) has equation \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \qquad a > b > 0\]
The line \(l\) is a normal to \(E\) at a point \(P(a\cos\theta, b\sin\theta)\), \(\ 0 < \theta < \dfrac{\pi}{2}\)
(a) Using calculus, show that an equation for \(l\) is \[ax\sin\theta - by\cos\theta = (a^2 - b^2)\sin\theta\cos\theta\] (5)
The line \(l\) meets the \(x\)-axis at \(A\) and the \(y\)-axis at \(B\).
(b) Show that the area of the triangle \(OAB\), where \(O\) is the origin, may be written as \(k\sin 2\theta\), giving the value of the constant \(k\) in terms of \(a\) and \(b\). (4)
(c) Find, in terms of \(a\) and \(b\), the exact coordinates of the point \(P\), for which the area of the triangle \(OAB\) is a maximum. (3)
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = -a\sin\theta\right.\) and \(\left.\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = b\cos\theta\right)\) so \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{b\cos\theta}{-a\sin\theta}\) M1: Differentiates both \(x\) and \(y\) and divides correctly A1: Fully correct derivative | M1 A1 |
| Normal has gradient \(\dfrac{a\sin\theta}{b\cos\theta}\) or \(\dfrac{a^2y}{b^2x}\) Correct perpendicular gradient rule | M1 |
| \((y - b\sin\theta) = \dfrac{a\sin\theta}{b\cos\theta}(x - a\cos\theta)\) Correct straight line method using a ‘changed’ gradient which is a function of \(\theta\) If \(y = mx + c\) is used need to find \(c\) for M1 | M1 |
| \(ax\sin\theta - by\cos\theta = (a^2 - b^2)\sin\theta\cos\theta\ ^*\) Fully correct completion to printed answer | A1 |
| (5) |
Notes
Alternative
| Scheme | Marks |
|---|---|
| M1: \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 \Rightarrow \dfrac{2x}{a^2} + \dfrac{2yy^{\prime}}{b^2} = 0 \Rightarrow y^{\prime} = -\dfrac{b^2x}{a^2y} = -\dfrac{b^2a\cos\theta}{a^2b\sin\theta}\) Differentiates implicitly and substitutes for \(x\) and \(y\) | M1 |
| A1: \(= -\dfrac{b\cos\theta}{a\sin\theta}\) | A1 |
| Scheme | Marks |
|---|---|
| \(x = \dfrac{(a^2 - b^2)\cos\theta}{a}\) Allow un-simplified | B1 |
| \(y = -\dfrac{(a^2 - b^2)\sin\theta}{b}\) Allow un-simplified | B1 |
| \(\left(= \tfrac{1}{2}\dfrac{(a^2 - b^2)^2\cos\theta\sin\theta}{ab}\right) = \tfrac{1}{4}\dfrac{(a^2 - b^2)^2}{ab}\sin 2\theta\) M1: Area of triangle is \(\tfrac{1}{2}\)"\(OA\)"\(\times\)"\(OB\)" and uses double angle formula correctly A1: Correct expression for the area (must be positive) | M1A1 |
| (4) |
| Scheme | Marks |
|---|---|
| Maximum area when \(\sin 2\theta = 1\) so \(\theta = \dfrac{\pi}{4}\) or 45 Correct value for \(\theta\) (may be implied by correct coordinates) | B1 |
| So the point \(P\) is at \(\left(\dfrac{a}{\sqrt{2}}, \dfrac{b}{\sqrt{2}}\right)\) oe \(\left(a\cos\dfrac{\pi}{4}, b\sin\dfrac{\pi}{4}\right)\) scores B1M1A0 M1: Substitutes their value of \(\theta\) where \(0 < \theta < \dfrac{\pi}{2}\) or \(0 < \theta < 90\) into their parametric coordinates A1: Correct exact coordinates | M1 A1 |
| Mark part (c) independently | |
| (3) | |
| (12 marks) |