Question

$4(x-y+2{)}^2+9(x+y-4{)}^2=25$ represents an ellipse whose :

• A. eccentricity is $\frac{5}{\sqrt{3}}$
• B. Length of major axis is $\frac{5}{\sqrt{2}}$
• C. Length of latus rectum is $\frac{10\sqrt{2}}{9}$
• D. Center is $(1,3)$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A eccentricity is $\frac{5}{\sqrt{3}}$
B Length of major axis is $\frac{5}{\sqrt{2}}$
C Length of latus rectum is $\frac{10\sqrt{2}}{9}$
D Center is $(1,3)$

$x-y+2=0$ and $x+y-4=0$ are perpendicular lines
Let
$\frac{x-y+2}{\sqrt{2}}=X$ and $\frac{x+y-4}{\sqrt{2}}=Y$

$4(\sqrt{2}X{)}^2+9(\sqrt{2}Y{)}^2=25$

$\Rightarrow \frac{X^2}{\frac{25}{8}}+\frac{Y^2}{\frac{25}{18}}=1$

$a^2=\frac{25}{8},b^2=\frac{25}{18}$
Center of ellipse is the point of intersection of $x-y+2=0$ and $x+y-4=0$ that is $(1,3)$

$e^2=1-\frac{b^2}{a^2}$ (as $a>b$)
$=1-\frac{8}{18}=\frac{10}{18}=\frac{5}{9}$
$e=\frac{\sqrt{5}}{3}$
Length of major axis is $2a=2\times \frac{5}{2\sqrt{2}}=\frac{5}{\sqrt{2}}$
Length of latus rectum
$=\frac{2b^2}{a}=2\times \frac{25}{18}\times \frac{2\sqrt{2}}{5}=\frac{10\sqrt{2}}{9}$