Question

What conics do the following equations represent? When possible, find their centres, and also their equations referred to the centre.
$13x^2-18xy+37y^2+2x+14y-2=0.$

Answer 1

The standard equation of any conic is written as $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$
When compared the given equation to this, we get
$a=13,h=-9,b=37,g=1,f=7,c=-2$
$h^2=81,ab=481$
$\therefore h^2<ab$
Also, $\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=\left|\begin{array}{ccc}13& -9& 1\\ -9& 37& 7\\ 1& 7& -2\end{array}\right|$
$=13\times (-74-49)+9\times (18-7)+1\times (-9\times 7-1\times 37)$
$=13\times (-123)+99-100$
$=-1599-1$
$=-1600$
Which is $<0$
When $h^2<ab$, the determinant is less than zero and $a$ and $b$ are both positive, the conic represents an ellipse.
Differentiating the conic equation w.r.t. $x$, we have
$26x-18y+2=0$ i.e. $13x-9y+1=0...\left(1\right)$
Differentiating the conic equation w.r.t. $y$, we have
$-18x+74y+14=0$ i.e. $-9x+37y+7=0...\left(2\right)$
Multiplying equation $\left(1\right)$ by $9$ and equation $\left(2\right)$ by $13$ and adding the two, we get
$-81y+9+481y+91=0$
$\therefore 400y=-100$ or $y=\frac{-1}{4}$
Correspondingly, $13x=9y-1=\frac{-9}{4}-1=\frac{-13}{4}$ or $x=\frac{-1}{4}$
The center of the ellipse is therefore $(\frac{-1}{4},\frac{-1}{4})$