Question

Which of the following option(s) is/are correct for ellipse $4(x-2y+1{)}^2+9(2x+y+2{)}^2=25$

• A. centre of ellipse is $(-1,0)$
• B. eccentricity $=\frac{\sqrt{5}}{3}$
• C. length of latus rectum is $\frac{4\sqrt{5}}{9}$ units
• D. ends of minor axis are
  $(-\frac{1}{3},\frac{1}{3})$ and $(-\frac{5}{3},-\frac{1}{3})$

Answer 1

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

The correct options are
A centre of ellipse is $(-1,0)$
B eccentricity $=\frac{\sqrt{5}}{3}$
C length of latus rectum is $\frac{4\sqrt{5}}{9}$ units
D ends of minor axis are

$(-\frac{1}{3},\frac{1}{3})$ and $(-\frac{5}{3},-\frac{1}{3})$
$4(x-2y+1{)}^2+9(2x+y+2{)}^2=25$
$\Rightarrow \frac{{\left(\frac{x-2y+1}{\sqrt{5}}\right)}^2}{{\left(\frac{\sqrt{5}}{2}\right)}^2}+\frac{{\left(\frac{2x+y+2}{\sqrt{5}}\right)}^2}{{\left(\frac{\sqrt{5}}{3}\right)}^2}=1$

Centre of the ellipse will be intersection of the lines $x-2y+1=0\left(\text{minor axis}\right)$ and $2x+y+2=0\left(\text{major axis}\right),$ which is $(-1,0).$
and

And $a=\frac{\sqrt{5}}{2},b=\frac{\sqrt{5}}{3}$
$\therefore e=\sqrt{1-\frac{b^2}{a^2}}=\frac{\sqrt{5}}{3}$

Length of Latus ractum $=\frac{2b^2}{a}=\frac{2{\left(\frac{\sqrt{5}}{3}\right)}^2}{\frac{\sqrt{5}}{2}}$
$=\frac{4\sqrt{5}}{9}$ units

minor axis $x-2y+1=0$ can be rewritten as
$\frac{x+1}{2}=y=r$
So any point on the line will be $P(2r-1,r)$
for end points of minor axis the distance of $P$ from major axis $=b$
$\therefore b=\frac{|4r-2+r+2|}{\sqrt{5}}\Rightarrow |5r|=\frac{5}{3}\Rightarrow r=\pm \frac{1}{3}$
Hence endpoint of minor axis are
$(-\frac{1}{3},\frac{1}{3})$ and $(-\frac{5}{3},-\frac{1}{3})$