Question

Find the foci of the curve $144x^2-120xy+25y^2+67x-42y+13=0$.

Answer 1

The foci are given by:-
$\frac{{(144x^{{}^{\prime }}-60y^{{}^{\prime }}+\frac{67}{2})}^2-{(-60x^{{}^{\prime }}+25y^{{}^{\prime }}-21)}^2}{119}$
$=\frac{\left(X\right)\left(Y\right)}{-60}$
$=144x^2-120xy+25y^2+67x-42y+13$
Where, $X=144x^{{}^{\prime }}-60y^{{}^{\prime }}+\frac{67}{2}$
$Y=60x^{{}^{\prime }}+25y^{{}^{\prime }}-21$
Simplifying the first pair gives
$60Y^2-119XY-60X^2=0$
$\Rightarrow (12Y+5X)(5Y-12X)=0$
$\Rightarrow 5Y-12X=0$
$\Rightarrow 12x^{{}^{\prime }}-5y^{{}^{\prime }}=-3$ ....(1)
$\Rightarrow X=-\frac{5}{2},Y=-6$
Putting in the second pair, we get
$\frac{-5.6}{2.60}=x^{{}^{\prime }}X+y^{{}^{\prime }}Y+\frac{67}{2}{x}^{{}^{\prime }}-21y^{{}^{\prime }}+13$
$224x^{{}^{\prime }}-108y^{{}^{\prime }}=-53$ ....(2)
Solving (1) and (2), we get
$x^{{}^{\prime }}=\frac{-59}{676},y^{{}^{\prime }}=\frac{66}{169}$
Co ordinates of focus are $(\frac{-59}{676},\frac{66}{169})$