Differentiation: YIJC 2024 Promo Q6

(a)

6(b)(i)$\dfrac{dx}{dt}=a(2t-1)$$\dfrac{dy}{dt}=a(-4t)$ $$\frac{dy}{dx}=\frac{-4t}{2t-1}$$ At $t=-1$, $\dfrac{dy}{dx}=\dfrac4{-3}$ $$\text{Gradient of normal is }\frac34$$ $$y-(-a)=\frac34(x-2a)$$ $$y=\frac34x-\frac52a$$ 6(b)(ii)$a(1-2t^2)=\dfrac34a(t^2-t)-\dfrac52a$ $$4-8t^2=3t^2-3t-10$$ $$11t^2-3t-14=0$$ $$(11t-14)(t+1)=0$$ $$t=\frac{14}{11}\text{ or }t=-1\quad\text{(rej. original point)}$$ $$x=a\left[\left(\frac{14}{11}\right)^2-\frac{14}{11}\right]=\frac{42}{121}a,$$ $$y=a\left[1-2\left(\frac{14}{11}\right)^2\right]=-\frac{271}{121}a$$ $$\left(\frac{42}{121}a,-\frac{271}{121}a\right).$$