Question

Trace the following central conics.
$2(3x-4y+5{)}^2-3(4x+3y-10{)}^2=150.$

Answer 1

Given equation of central conics is

$2{(3x-4y+5)}^2-3{(4x+3y-10)}^2=150$

$2\times 25{\left(\frac{3x-4y+5}{5}\right)}^2-3\times 25{\left(\frac{3x-4y+5}{5}\right)}^2=150$

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

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

$\Rightarrow \frac{X^2}{3}-\frac{Y^2}{2}=1$

Comparing with the standard equation of hyperbola, we get

$a=\sqrt{3},b=\sqrt{2}$