Question

What conics do the following equations represent? When possible, find their centres, and also their equations referred to the centre.
$2x^2-72xy+23y^2-4x-28y-48=0.$

Answer 1

Given conic $S:2x^2-72xy+23y^2-4x-28y-48=0$

The general equation of a conic section is a second-degree can be written as

$f(x,y)=a{x}^2+2hxy+b{y}^2+2gx+2fy+c$

Here, $h=-36,a=2,b=23,g=-2,f=-14,c=-48$

$△=abc+2fgh-a{f}^2-b{g}^2-c{h}^2$

$=\left(2\right)\left(23\right)(-48)+2(-14)(-2)(-36)-2(-14{)}^2-23(-2{)}^2-(-48)(-36{)}^2$

$=-2208-2016-392-92+62208$

$=59300\ne 0$

$h^2-ab=(-36{)}^2-2\times 23=1250$

i.e. $h^2-ab>0$

So, it represents a hyperbola.

Now, $\frac{\partial S}{\partial x}=4x-72y-4$

$\frac{\partial S}{\partial y}=46y-72x-28$

For centre, $\frac{\partial S}{\partial x}=0$ and $\frac{\partial S}{\partial y}=0$

So, $4x-72y-4=0$ $.......................\left(i\right)$

and $46y-72x-28=0$ $......................\left(ii\right)$

Solving eqn. $\left(i\right)$ and $\left(ii\right)$, we get

$x=-\frac{11}{25}$ and $y=-\frac{2}{25}$

$\therefore C(-\frac{11}{25},-\frac{2}{25})$

Now,

$C^{\prime }=gx+fy+c=-2\times (-\frac{11}{25})-14\times (-\frac{2}{25})-48$

$=\frac{22}{25}+\frac{28}{25}-48$

$=2-48$

$=-46$