Question

Trace the parabolas :
$16x^2-24xy+9y^2-5x-10y+1=0$.

Answer 1

$16x^2-24xy+9y^2-5x-10y+1=0$

${(4x-3y)}^2=5x+10y-1$

Introducing a variable $\lambda$

${(4x-3y+\lambda )}^2=5x+10y-1-6\lambda y+8\lambda x+{\lambda }^2$

${(4x-3y+\lambda )}^2=(5+8\lambda )x+(10-6\lambda )y+{\lambda }^2-1$

$4x-3y+\lambda =\mathrm{0.......}\left(i\right)$

$(5+8\lambda )x+(10-6\lambda )y+{\lambda }^2=\mathrm{0.......}\left(ii\right)$

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

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

$m=-\frac{4}{-3}=\frac{4}{3}{m}^{\prime }=-\frac{5+8\lambda }{10-6\lambda }$

$m{m}^{\prime }=-1$

$\frac{4}{3}\times -\frac{5+8\lambda }{10-6\lambda }=-1$

$20+32\lambda =30-18\lambda$

$\Rightarrow \lambda =\frac{1}{5}$

${(4x-3y+\frac{1}{5})}^2=(5+8.\frac{1}{5})x+(10-6.\frac{1}{5})y+{\frac{1}{5}}^2-1$

${(4x-3y+\frac{1}{5})}^2=\frac{33}{5}x+\frac{44}{5}y-\frac{24}{25}$

${(4x-3y+\frac{1}{5})}^2=\frac{11}{5}(3x+4y-\frac{24}{55})$

${\left(\frac{4x-3y+\frac{1}{5}}{5}\right)}^2=\frac{11}{25}\left(\frac{3x+4y-\frac{24}{55}}{5}\right)$

$Y=\frac{4x-3y+\frac{1}{5}}{5},X=\frac{3x+4y-\frac{24}{55}}{5}$

$Y^2=\frac{11}{25}X$

Comparing with standard equation of parabola

$4a=\frac{11}{25}$

$\Rightarrow a=\frac{11}{100}$