2010 H2 Mathematics Paper 1 Question 11

Differentiation I: Tangents and Normals, Parametric Curves

Answers

(ii)

Area of △OAB=4{\triangle OAB = 4} which is independent of p{p}

(iii)

x2−y2=4{x^2-y^2=4}

Full solutions

(i)

dxdt=1−1t2=t2−1t2dydt=1+1t2=t2+1t2\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} &= 1 - \frac{1}{t^{2}} \\ &= \frac{ t^2 - 1 }{ t^2 } \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= 1 + \frac{1}{t^{2}} \\ &= \frac{ t^2 + 1 }{ t^2 } \end{align*}
dydx=dydt÷dxdt=t2+1t2÷t2−1t2=t2+1t2−1\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t} \\ &=\frac{ t^2 + 1 }{ t^2 } \div \frac{ t^2 - 1 }{ t^2 } \\ &=\frac{ t^2 + 1 }{ t^2 - 1 } \\ \end{align*}
At point P{P} with parameter p,{p,}
dydx=p2+1p2−1\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{ p^2 + 1 }{ p^2 - 1 }
Tangent at P:{P:}
y−(p−1p)=p2+1p2−1(x−(p+1p))(p2−1)(y−p2−1p)=(p2+1)(x−p2+1p)(p2−1)y−p4−2p2+1p=(p2+1)x−p4+2p2+1p(p2+1)x−(p2−1)y=p4+2p2+1−(p4−2p2+1)p\begin{gather*} y - \left( p - \frac{1}{p} \right) = \frac{ p^2 + 1 }{ p^2 - 1 } \left( x - \left( p + \frac{1}{p} \right) \right) \\ (p^2-1) \left ( y - \frac{p^2-1}{p} \right) = (p^2+1) \left( x - \frac{p^2+1}{p} \right) \\ (p^2-1) y - \frac{p^4 - 2p^2 + 1}{p} = (p^2+1) x - \frac{p^4 + 2p^2 + 1}{p} \\ (p^2+1)x - (p^2-1)y = \frac{p^4+2p^2+1-(p^4-2p^2+1)}{p} \\ \end{gather*}
Hence equation of tangent at P{P} is
(p2+1)x−(p2−1)y=4p  ■(p^2+1)x - (p^2-1)y = 4p \; \blacksquare

(ii)

At point A,{A, } substituting y=x{y=x} into equation of tangent at P,{P,}
(p2+1)x−(p2−1)x=4p2x=4px=2p\begin{gather*} (p^2+1)x - (p^2-1)x = 4p \\ 2x = 4p \\ x = 2p \end{gather*}
At point B,{B, } substituting y=−x{y=-x} into equation of tangent at P,{P,}
(p2+1)x+(p2−1)x=4p2p2x=4px=2p\begin{gather*} (p^2+1)x + (p^2-1)x = 4p \\ 2p^2 x = 4p \\ x = \frac{2}{p} \end{gather*}
Hence coordinates of A{A} and B{B} are A(2p,2p){A(2p,2p)} and B(2p,−2p){B \left(\frac{2}{p}, -\frac{2}{p} \right)}
As the lines y=x{y=x} and y=−x{y=-x} are perpendicular,
Area of △OAB=12(OA×OB)=12(2p)2+(−2p)2(2p)2+(−2p)2=128p28p2=12(8p)(8p)=4 units2\begin{align*} & \textrm{Area of } \triangle OAB \\ &= \frac{1}{2} (OA \times OB) \\ &= \frac{1}{2} \sqrt{(2p)^2+(-2p)^2} \sqrt{\left(\frac{2}{p}\right)^2+\left(-\frac{2}{p}\right)^2} \\ &= \frac{1}{2} \sqrt{8p^2} \sqrt{\frac{8}{p}^2} \\ &= \frac{1}{2} (\sqrt{8}p) \left( \frac{\sqrt{8}}{p} \right) \\ &= 4 \textrm{ units}^2 \\ \end{align*}
Hence the area of △OAB{\triangle OAB} is independent of p  ■{p \; \blacksquare}

(iii)

x=t+1ty=t−1t\begin{align} && \quad x = t + \frac{1}{t} \\ && \quad y = t - \frac{1}{t} \\ \end{align}
Taking (1)+(2),{(1)+(2),}
x+y=2t\begin{equation} x+y = 2t \end{equation}
Taking (1)−(2),{(1)-(2),}
x−y=2t\begin{equation} x-y = \frac{2}{t} \end{equation}
Equating t{t} from (3){(3)} and (4),{(4),}
x+y2=2x−y\frac{x+y}{2} = \frac{2}{x-y}
Hence the cartesian equation of C:{C: }
x2−y2=4  ■x^2 - y^2 = 4 \; \blacksquare
This is a hyperbola with centered at the origin with asymptotes y=±x{y=\pm x}