Question

Trace the parabolas :
$9x^2+24xy+16y^2-4y-x+7=0.$

Answer 1

${(3x+4y)}^2=x+4y-7$

Introducing the variable $\lambda$

${(3x+4y+\lambda )}^2=x+4y+8\lambda y+6\lambda x+{\lambda }^2-7{(3x+4y+\lambda )}^2=(1+6\lambda )x+(4+8\lambda )y+{\lambda }^2-73x+4y+\lambda =\mathrm{0......}\left(i\right)(1+6\lambda )x+(4+8\lambda )y+{\lambda }^2-7=\mathrm{0......}\left(ii\right)$

For $\lambda$ we assume both the lines are perpendicular

Let slope of $\left(i\right)$ be $m$ and $\left(ii\right)$ be $m^{\prime }$

$m=-\frac{3}{4}{m}^{\prime }=-\frac{1+6\lambda }{4+8\lambda }m{m}^{\prime }=-1-\frac{3}{4}\times -\frac{1+6\lambda }{4+8\lambda }=-13+18\lambda =-16-32\lambda \Rightarrow \lambda =-\frac{19}{50}{(3x+4y+\lambda )}^2=(1+6\lambda )x+(4+8\lambda )y+{\lambda }^2-7{(3x+4y-\frac{19}{50})}^2=(1+6\times -\frac{19}{50})x+(4+8\times -\frac{19}{50})+{(-\frac{19}{50})}^2-7{(3x+4y-\frac{19}{50})}^2=\frac{-32}{25}x+\frac{24}{25}y-\frac{17139}{2500}{(3x+4y-\frac{19}{50})}^2=-\frac{8}{25}(4x-3y+\frac{17139}{800}){\left(\frac{3x+4y-\frac{19}{50}}{5}\right)}^2=-\frac{8}{125}\left(\frac{4x-3y+\frac{17139}{800}}{5}\right)Y=3x+4y-\frac{19}{50},X=\frac{4x-3y+\frac{17139}{800}}{5}{Y}^2=-\frac{8}{125}X$

Comparing with the standard equation of parabola

$4a=-\frac{8}{125}\Rightarrow a=-\frac{2}{125}$