A2 October 2020 Q5

Notes

M1: Uses \(b^2 = a^2(1 - e^2)\) with \(a = 6\) and \(b = 4\) to find a value for \(e^2\) or \(e\).

A1: Correct value for \(e\) or \(e^2\)

A1: Correct foci

Notes

M1: Forms a complete strategy to find the perimeter using general \(P\) and applies the focus directrix property to the sides \(PS\) and \(PS^{\prime}\)

M1: Deduces the length of the two sides adjacent to \(P\) is a constant

B1ft: Uses \(SS^{\prime}\) is twice their \(ae\) from (a)

A1*: Finds the value and makes conclusion that perimeter is constant for any \(P\) on \(E\)

Alternative

Scheme	Marks	AO
Perimeter \(= PS + PS^{\prime} + SS^{\prime}\) \(PS = \sqrt{\left(2\sqrt{5} - 6\cos\theta\right)^2 + (4\sin\theta)^2} = \ldots\left\{6 - 2\sqrt{5}\cos\theta\right\}\) \(PS^{\prime} = \sqrt{\left(2\sqrt{5} + 6\cos\theta\right)^2 + (4\sin\theta)^2} = \ldots\left\{6 + 2\sqrt{5}\cos\theta\right\}\)	M1	3.1a
\(PS + PS^{\prime} = \left(6 - 2\sqrt{5}\cos\theta\right) + \left(6 + 2\sqrt{5}\cos\theta\right) = 12\)	M1	2.2a
\(\ldots + 2 \times 2\sqrt{5}\)	B1ft	1.1b
\(= 12 + 4\sqrt{5}\) hence perimeter is constant for any \(P\) on \(E\). *	A1*	2.1
	(4)

M1: Forms a complete strategy to find the perimeter using general \(P\). Finds the lengths of \(PS\) and \(PS^{\prime}\) using Pythagoras theorem and the general coordinate \((6\cos\theta, 4\sin\theta)\)

M1: Deduces the length of the two sides adjacent to \(P\) is a constant, using trig identities.

B1ft: Uses \(SS^{\prime}\) is twice their \(ae\) from (a)

A1*: Finds the value and makes conclusion that perimeter is constant for any \(P\) on \(E\)

(Corrected from the printed mark scheme: the second line of the alternative ends “\(= B\)”; the sum is \(12\).)

Note: Using the property that \(PS + PS^{\prime} = 2a\) both method marks may be awarded as long as a reason is given e.g. definition/property of an ellipse