FP1 June 2018 Q5

EdexcelOld spec9 marksConic Sections

5. The rectangular hyperbola \(H\) has equation \(xy = c^2\), where \(c\) is a positive constant.

Given that \(P\left(ct, \dfrac{c}{t}\right)\), \(t \neq 0\), is a general point on \(H\),

(a) use calculus to show that the equation of the tangent to \(H\) at \(P\) can be written as \[t^2y + x = 2ct\] (4)

The points \(A\) and \(B\) lie on \(H\).

The tangent to \(H\) at \(A\) and the tangent to \(H\) at \(B\) meet at the point \(\left(-\dfrac{8c}{5}, \dfrac{3c}{5}\right)\).

Given that the \(x\) coordinate of \(A\) is positive,

(b) find, in terms of \(c\), the coordinates of \(A\) and the coordinates of \(B\). (5)

Mark scheme (a)

Scheme	Marks
\(y = c^2x^{-1} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -c^2x^{-2}\) or (implicitly) \(y + x\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) or (chain rule) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -ct^{-2} \times \dfrac{1}{c}\)	M1
When \(x = ct\), \(m_T = \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-c^2}{(ct)^2} = -\dfrac{1}{t^2}\) or at \(P\left(ct, \dfrac{c}{t}\right)\), \(m_T = \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{y}{x} = -\dfrac{ct^{-1}}{ct} = -\dfrac{1}{t^2}\)	A1
T: \(y - \dfrac{c}{t} = -\dfrac{1}{t^2}(x - ct)\)	M1
T: \(t^2y - ct = -x + ct\)
T: \(t^2y + x = 2ct\quad *\)	A1 cso *
	(4)

Notes

M1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \pm kx^{-2}\) or \(y + x\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) or \(\dfrac{\text{their } \frac{\mathrm{d}y}{\mathrm{d}t}}{\text{their } \frac{\mathrm{d}x}{\mathrm{d}t}}\)

A1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{t^2}\)

M1: Applies \(y - \dfrac{c}{t} = \left(\text{their } m_T\right)(x - ct)\) where their \(m_T\) has come from calculus

At least one line of working.

A1 cso *: Correct solution.

Mark scheme (b)

Scheme	Marks
\(t^2\left(\dfrac{3c}{5}\right) + \left(-\dfrac{8c}{5}\right) = 2ct\)	M1
\(3t^2 - 8 = 10t\)	A1
\(\left\{3t^2 - 10t - 8 = 0 \Rightarrow\right\}\ (t - 4)(3t + 2) = 0 \Rightarrow t = \ldots\)	M1
\(t = 4,\ -\dfrac{2}{3} \Rightarrow A\left(4c, \dfrac{c}{4}\right),\ B\left(-\dfrac{2}{3}c, -\dfrac{3c}{2}\right)\)	M1 A1
	(5)
	(9 marks)

Notes

M1: Substitutes \(\left(-\dfrac{8c}{5}, \dfrac{3c}{5}\right)\) into tangent.

A1: Correct 3TQ in terms of \(t\). Can include uncancelled \(c\).

M1: Attempt to solve their 3TQ for \(t\)

M1: Uses one of their values of \(t\) to find \(A\) or \(B\)

A1: Correct coordinates. Condone \(A\) and \(B\) swapped or missing.

ALT 1 (b)

Scheme	Marks
\(y - \dfrac{3c}{5} = -\dfrac{1}{t^2}\left(x - -\dfrac{8c}{5}\right)\) \(\Rightarrow \dfrac{c}{t} - \dfrac{3c}{5} = -\dfrac{1}{t^2}\left(ct + \dfrac{8c}{5}\right)\)	M1
\(3t^2 - 10t = 8\)	A1
then apply the original mark scheme.

M1: Substitutes \(\left(ct, \dfrac{c}{t}\right)\) into their \(y - \dfrac{3c}{5} = -\dfrac{1}{t^2}\left(x - -\dfrac{8c}{5}\right)\)

A1: Correct 3TQ in terms of \(t\). Can include uncancelled \(c\).

ALT 2 (b)

Scheme	Marks
\(A\left(ct_1, \dfrac{c}{t_1}\right),\ B\left(ct_2, \dfrac{c}{t_2}\right)\) \({t_1}^2y + x = 2ct_1\) \({t_2}^2y + x = 2ct_2\)	M1
\(t_1 + t_2 = \dfrac{10}{3},\ t_1t_2 = -\dfrac{8}{3}\)
\(3t^2 - 8 = 10t\)	A1
then apply original scheme

M1: Substitutes \(A\) and \(B\) into the equation of the tangent, solves for \(x\) and \(y\)

A1: Correct 3TQ in terms of \(t_1\) or \(t_2\). Can include uncancelled \(c\).