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Q.
The equation $9 x^{2}-16 y^{2}-18 x+32 y-151=0$ represents a hyperbola
Conic Sections
Solution:
We have, $9 x^{2}-16 y^{2}-18 x+32 y-151=0$
$\Rightarrow 9\left(x^{2}-2 x\right)-16\left(y^{2}-2 y\right)=151$
$\Rightarrow 9\left(x^{2}-2 x+1\right)-16\left(y^{2}-2 y+1\right)=144$
$\Rightarrow 9(x-1)^{2}-16(y-1)^{2}=144$
$\Rightarrow \frac{(x-1)^{2}}{16}-\frac{(y-1)^{2}}{9}=1$
Shifting the origin at $(1,1)$ without rotating the axes
$\frac{ X ^{2}}{16}-\frac{ Y ^{2}}{9}=1$, where $x = X +1$ and $y = Y +1$
This is of the form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
where $a^{2}=16$ and $b^{2}=9$
so the length of the transverse axes $=2 a =8$
The length of the latus rectum $=\frac{2 b ^{2}}{ a }=\frac{ a }{2}$
The equation of the directrix, $x=\pm \frac{a}{c}$
$x -1=\pm \frac{16}{5} $
$\Rightarrow x =\pm \frac{16}{5}+1 $
$ \Rightarrow x =\frac{21}{5} ; x =-\frac{11}{5}$