Q.(a) A missile fired from ground level rises $x$ metres vertically upwards in $t$ seconds and $x = 100t - \dfrac{25}{2}t^2$. Find:

(a) differentiates the given height function to get velocity, then reads off the initial velocity, time/height of the peak, and the striking velocity; (b) completes the square twice to bring the given hyperbola to standard form and reads off its centre, foci and vertices.

(a) Missile kinematics, $x=100t-\dfrac{25}{2}t^2$

1. Velocity: $v(t)=\dfrac{dx}{dt}=100-25t$.
2. (i) Initial velocity: $v(0)=100$ m/s.
3. (ii) Time of maximum height: at the peak $v=0\Rightarrow 100-25t=0\Rightarrow t=4$ s (and $\dfrac{d^2x}{dt^2}=-25<0$ confirms a maximum).
4. (iii) Maximum height: $x(4)=100(4)-\dfrac{25}{2}(4)^2=400-200=200$ m.
5. (iv) Striking velocity: the missile returns to ground level ($x=0$) when $100t-\dfrac{25}{2}t^2=0\Rightarrow t\Big(100-\dfrac{25}{2}t\Big)=0\Rightarrow t=0$ or $t=8$ s (landing time). Then $v(8)=100-25(8)=100-200=-100$ m/s, so it strikes the ground with speed $100$ m/s (directed downward).

(b) Hyperbola $16x^2-9y^2-32x-18y+151=0$

1. Group and complete the square: $16(x^2-2x)-9(y^2+2y)+151=0$.
2. $16\big[(x-1)^2-1\big]-9\big[(y+1)^2-1\big]+151=0 \Rightarrow 16(x-1)^2-16-9(y+1)^2+9+151=0$.
3. $16(x-1)^2-9(y+1)^2+144=0 \Rightarrow 16(x-1)^2-9(y+1)^2=-144$.
4. Divide by $-144$: $-\dfrac{(x-1)^2}{9}+\dfrac{(y+1)^2}{16}=1$, i.e. $\dfrac{(y+1)^2}{16}-\dfrac{(x-1)^2}{9}=1$ — a hyperbola with vertical transverse axis.
5. Standard form $\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$: centre $(h,k)=(1,-1)$, $a^2=16$ ($a=4$), $b^2=9$ ($b=3$), and $c^2=a^2+b^2=25\Rightarrow c=5$.
6. Vertices ($h,k\pm a$): $(1,-1+4)=(1,3)$ and $(1,-1-4)=(1,-5)$. …