Question

The equation $9x^2-16y^2-18x+32y-151=0$ represents a hyperbola

• A. The length of the transverse axes is 4
• B. Length of latus rectum is 9
• C. Equation of directrix is $x=\frac{21}{5}$ and $x=-\frac{11}{5}$
• D. None of these

Answer 1

Answer: (C)

We have, $9x^2-16y^2-18x+32y-151=0$
$\Rightarrow 9\left(x^2-2x\right)-16\left(y^2-2y\right)=151$
$\Rightarrow 9\left(x^2-2x+1\right)-16\left(y^2-2y+1\right)=144$
$\Rightarrow 9(x-1{)}^2-16(y-1{)}^2=144$
$\Rightarrow \frac{(x-1{)}^2}{16}-\frac{(y-1{)}^2}{9}=1$
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 $=2a=8$
The length of the latus rectum $=\frac{2b^2}{a}=\frac{a}{2}$
The equation of the directrix, $x=\pm \frac{a}{c}$
$x-1=\pm \frac{16}{5}$
$\Rightarrow x=\pm \frac{16}{5}+1$
$\Rightarrow x=\frac{21}{5};x=-\frac{11}{5}$