Question

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

• A. $10$
• B. $\frac{20}{3}$
• C. $\frac{20}{7}$
• D. $4$

Answer 1

The correct option is B $\frac{20}{3}$

$(5x-10{)}^2+(5y+15{)}^2=\frac{(3x-4y+7{)}^2}{4}$
$\Rightarrow (x-2{)}^2+(y+3{)}^2={\left(\frac{1}{2}\frac{3x-4y-7}{5}\right)}^2$
$\Rightarrow \sqrt{(x-2{)}^2+(y+3{)}^2}=\frac{1}{2}\frac{|3x-4y-7|}{5}$
It is an ellipse, whose focus is (2, -3),
Directrix is $3x-4y+7=0$,
Eccentricity $\frac{1}{2}$.
Length of perpendicular from the focus to the directrix is
$\frac{|3\times 2-4(-3)+7|}{5}=5$
$\Rightarrow \frac{a}{e}-ae=5$
$\Rightarrow 2a-\frac{a}{2}=5\Rightarrow a=\frac{10}{3}$
So, the length of the major axis is $\frac{20}{3}$.