Question

If the length of the major axis of the ellipse $(5x-10{)}^2+(5y+15{)}^2=\frac{(3x-4y+7{)}^2}{4},$ is $k$ units, then $3k=$

Answer 1

$(5x-10{)}^2+(5y+15{)}^2=\frac{(3x-4y+7{)}^2}{4}\Rightarrow (x-2{)}^2+(y+3{)}^2={(\frac{1}{2}\times \frac{3x-4y+7}{5})}^2\Rightarrow \sqrt{(x-2{)}^2+(y+3{)}^2}=\frac{1}{2}\times \frac{|3x-4y+7|}{5}$
This represents an ellipse, whose focus is $(2,-3)$, directrix is $3x-4y+7=0$ and eccentricity is $\frac{1}{2}$.

Length of perpendicular from focus to directrix is
$=\frac{|3\times 2-4(-3)+7|}{5}=5$ units
Also,
$\frac{a}{e}-ae=5\Rightarrow 2a-\frac{a}{2}=5\Rightarrow a=\frac{10}{3}$

Hence, the length of major axis is $\frac{20}{3}\text{units}$
$\therefore 3k=20$