June Revision Exercise 3 Q2

By KS Teng

(i)

\frac{dx}{dt}=-\frac{3a}{t^4}

\frac{dy}{dt}=-\frac{a}{t^2}

\frac{dy}{dx}= \frac{dy}{dt} \times \frac{dt}{dx} = \frac{t^2}{3}

When t= \frac{1}{2}, \frac{dy}{dx}= \frac{1}{12}

Equation of tangent: y-2a = \frac{1}{12}(x-8a) \Rightarrow y=\frac{1}{12}x + \frac{4}{3}a

Equation of normal: y-2a = -12(x-8a) \Rightarrow y=-12x+98a

(ii)
\frac{a}{t}=\frac{1}{12}\frac{a}{t^3}+\frac{4}{3}a

16t^3 - 12t^2 + 1 =0

t=\frac{1}{2} \text{ (rej) or } t=-\frac{1}{4}

\text{When } t=-\frac{1}{4}, x= -64a, y=-4a

Hence, tangent cuts curve again at (-64a, -4a)

(iii)
At Q, y=0 \Rightarrow x=-16a

At R, y=0 \Rightarrow x=\frac{49}{6}a

\text{Area } = \frac{1}{2}[\frac{49}{6}a-(-16a)](2a) = \frac{145}{6}a^2 \text{units}^2

Back to June Revision Exercise 3

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June Revision Exercise 3 Q1JC Mathematics
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