Question

Trace the parabolas $144x^2-120xy+25y^2+619x-272y+663=0,$ and find its focus.

Answer 1

$44x^2-120xy+25y^2+619x-272y+663=0{(12x-5y)}^2=272y-619x-663$

Introducing a variable $\lambda$

${(12x-5y+\lambda )}^2=272y-619x-663-10\lambda y+24\lambda x+{\lambda }^2{(12x-5y+\lambda )}^2=(272-10\lambda )y-(619-24\lambda )x+{\lambda }^2-663$

$12x-5y+\lambda =0......\left(i\right)(272-10\lambda )y-(619-24\lambda )x+{\lambda }^2-663=\mathrm{0......}\left(ii\right)$

For $\lambda$ we assume both lines to be perpendicular

$m=-\frac{12}{-5}=\frac{12}{5}$

$m^{\prime }=-\frac{-(619-24\lambda )}{(272-10\lambda )}=\frac{(619-24\lambda )}{(272-10\lambda )}$

$m{m}^{\prime }=-1\frac{12}{5}\times \frac{(619-24\lambda )}{(272-10\lambda )}=-17428-288\lambda =50\lambda -1360\Rightarrow \lambda =26{(12x-5y+26)}^2=(272-10\times 26)y-(619-24\times 26)x+{\left(26\right)}^2-663{(12x-5y+26)}^2=5x+12y+13{\left(\frac{12x-5y+26}{13}\right)}^2=\frac{1}{13}\left(\frac{5x+12y+13}{13}\right)Y=\frac{12x-5y+26}{13},X=\frac{5x+12y+13}{13}{Y}^2=\frac{1}{13}X$

Comparing with the standard equation of parabola $y^2=4ax$, we get

$4a=\frac{1}{13}a=\frac{1}{52}$

The focus of the parabola is obtained by solving $X-a=0$ and $Y=0$

$\frac{5x+12y+13}{13}-a=0\frac{5x+12y+13}{13}-\frac{1}{52}=020x+48y+51=\mathrm{0........}\left(iii\right)\frac{12x-5y+26}{13}=012x-5y+26=\mathrm{0........}\left(iv\right)$

Solving $\left(iii\right)$ and $\left(iv\right)$

we get $x=-\frac{1503}{676}$ and $y=-\frac{23}{169}$

So the focus is $(-\frac{1503}{676},-\frac{23}{169})$