Question

Find the products of the semi-axes of the conics $4(3x+4y-7{)}^2+3(4x-3y+9{)}^2=3.$

Answer 1

${4(3x+4y-7)}^2+3{(4x-3y+9)}^2=3100\frac{{(3x+4y-7)}^2}{25}+75\frac{{(4x-3y+9)}^2}{25}=3100{\left(\frac{3x+4y-7}{5}\right)}^2+75{\left(\frac{4x-3y+9}{5}\right)}^2=3\frac{100}{3}{\left(\frac{3x+4y-7}{5}\right)}^2+25{\left(\frac{4x-3y+9}{5}\right)}^2=1$

Let $\frac{3x+4y-7}{5}=X$ and $\frac{4x-3y+9}{5}=Y$

$\frac{100}{3}{X}^2+25Y^2=1\frac{X^2}{{\left(\frac{\sqrt{3}}{10}\right)}^2}+\frac{Y^2}{{\left(\frac{1}{5}\right)}^2}=1$

Comparing with the standard equation of ellipse, we get

$a=\frac{\sqrt{3}}{10},b=\frac{1}{5}$

Product of semi axes $=ab$$=\frac{\sqrt{3}}{10}\times \frac{1}{5}=\frac{\sqrt{3}}{50}$