focus is (2, 2), directive is x + y = 9 and eccentricity = 2.

Question

focus is $(2, 2)$, directive is $x + y = 9$ and eccentricity = $2$.

✓

Answer

Let $\mathrm{S}(2,2)$ be the focus and $\mathrm{P}(\mathrm{x}, \mathrm{y})$ be a point on the hyperbola. Draw PM perpendicular from P on the directrix. Then, by definition $s P=e P M \Rightarrow \mathrm{sP}^2=\mathrm{e}^2 \mathrm{PM}^2 \Rightarrow(\mathrm{x}-2)^2+(\mathrm{y}-2)^2=2^2\left[\frac{\mathrm{x}+\mathrm{y}-9}{\sqrt{1^2+1^2}}\right]\left[\because \mathrm{e}=\frac{4}{3}\right]$ $\Rightarrow \mathrm{x}^2+4-4 \mathrm{x}+\mathrm{y}^2+4-4 \mathrm{y}=\frac{4[\mathrm{x}+\mathrm{y}-9]^2}{2}$
$\Rightarrow \mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-4 \mathrm{y}+8=2[\mathrm{x}+\mathrm{y}-9]^2$
$\Rightarrow \mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-4 \mathrm{y}+8=2[\mathrm{x}+\mathrm{y}+(-9)+2 \times \mathrm{x} \times \mathrm{y}+2 \times \mathrm{y} \times(-9)+2 \times(-9) \times \mathrm{x}]$
$\Rightarrow \mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-4 \mathrm{y}+8=2\left[\mathrm{x}^2+\mathrm{y}^2+81+2 \mathrm{xy}-18 \mathrm{y}+18 \mathrm{x}\right]$
$\Rightarrow \mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-4 \mathrm{y}+8=\left[2 \mathrm{x}^2+2 \mathrm{y}^2+162+4 \mathrm{xy}-36 \mathrm{y}+36 \mathrm{x}\right] $
$\Rightarrow 2 \mathrm{x}^2-\mathrm{x}^2+2 \mathrm{y}^2-\mathrm{y}^2+4 \mathrm{xy}-36 \mathrm{x}+4 \mathrm{x}-36 \mathrm{y}+4 \mathrm{y}+162-8=0$
$\Rightarrow \mathrm{x}^2+y^2-4 x y-32 x-32 y+154=0$
This is the required equation of the hyperbola.