June Revision Exercise 3 Q2




\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


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)

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

Leave a Comment

thirteen + 20 =

Contact Us

CONTACT US We would love to hear from you. Contact us, or simply hit our personal page for more contact information

Not readable? Change text. captcha txt

Start typing and press Enter to search