Q.Find the equation of the hyperbola if its centre is $(2, 1)$; one of the foci is $(8, 1)$ and the corresponding directrix is $x=4$. OR

Main part: use the centre-focus-directrix relations for a hyperbola to find $a^2,b^2$. OR alternative: simplify $\frac{1+i}{1-i}$ to a power of $i$, then use De Moivre's theorem for the cube roots of $i$.

Main part — equation of the hyperbola

1. Given data. Centre $C=(2,1)$; focus $F=(8,1)$; corresponding directrix $x=4$. Since $C$ and $F$ share the same $y$-coordinate, the transverse axis is horizontal.

2. Find $c$ (centre-to-focus distance).

$c=|8-2|=6$

3. Use the centre-to-directrix relation. For a horizontal hyperbola, the directrix corresponding to a focus on the same side lies at distance $\dfrac{a^2}{c}=\dfrac{a}{e}$ from the centre (since $e=c/a$):

$\left|4-2\right|=\frac{a^2}{c} \ \Rightarrow\ 2=\frac{a^2}{6} \ \Rightarrow\ a^2=12$

4. Find $b^2$ using $b^2=c^2-a^2$ (hyperbola relation).

$b^2=6^2-12=36-12=24$

5. Write the equation with centre $(h,k)=(2,1)$:

$\frac{(x-2)^2}{12}-\frac{(y-1)^2}{24}=1$

OR — alternative part

(i) Least positive integer $n$ with $\left(\dfrac{1+i}{1-i}\right)^n=1$

Step 1: Simplify $\dfrac{1+i}{1-i}$. Multiply numerator and denominator by the conjugate $(1+i)$:

$\frac{1+i}{1-i}\cdot\frac{1+i}{1+i}=\frac{(1+i)^2}{1-i^2}=\frac{1+2i-1}{1+1}=\frac{2i}{2}=i$

Step 2: Solve $i^n=1$. The powers of $i$ cycle with period 4: $i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1$. The least positive integer for which $i^n=1$ is:

$n=4$

(ii) Cube roots of $i$, i.e. $i^{1/3}$

Step 1: Write $i$ in polar (trigonometric) form.

$i=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}$

Step 2: General polar form with period $2k\pi$.

$i=\cos\left(\frac{\pi}{2}+2k\pi\right)+i\sin\left(\frac{\pi}{2}+2k\pi\right),\qquad k=0,1,2$

Step 3: Apply De Moivre's theorem for the cube root. …