Question

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

• A. $\frac{20}{3}$
• B. $\frac{15}{3}$
• C. $\frac{10}{3}$
• D. $\frac{25}{3}$

Answer 1

Answer: (A)

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