The length of the major axis of the ellipse $(5\mathrm{x}-10{)}^2+(5\mathrm{y}+15{)}^2=\frac{(3\mathrm{x}-4\mathrm{y}+7{)}^2}{4} \mathrm{is}$

$\begin{array}{l}(5\mathrm{x}-10{)}^2+(5\mathrm{y}+15{)}^2=\frac{(3\mathrm{x}-4\mathrm{y}+7{)}^2}{4}\\ \text{⇒ }(\mathrm{x}-2{)}^2+(\mathrm{y}+3{)}^2={\left(\frac{1}{2}\frac{3\mathrm{x}-4\mathrm{y}+7}{5}\right)}^2\\ \text{⇒ }\sqrt{(\mathrm{x}-2{)}^2+(\mathrm{y}+3{)}^2}=\frac{1}{2}\frac{\mid 3\mathrm{x}-4\mathrm{y}+7}{5}\end{array}$
It is an ellipse, whose focus is (2, -3), directrix is 3x - 4y + 7 = 0, and eccentricity is 1/2.
Length of perpendicular from the focus to the directrix is
$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{|3\times 2-4(-3)+7|}{5}=5\\ \text{⇒ }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{\mathrm{a}}{\mathrm{e}}-\mathrm{ae}=5\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\mathrm{a}-\frac{\mathrm{a}}{2}=5\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{a}=\frac{10}{3}\end{array}$
So, the length of the major axis is 20/3.