Question

For the given ellipse $9x^2+25y^2-18x-100y-116=0$, which of the following is/are correct?

• A. centre is $(1,2)$
• B. eccentricity is $\frac{4}{5}$
• C. end points of latus-rectum are $(5,\frac{19}{5})\text{and}(5,\frac{1}{5})$
• D. length of major axis will be $10$ units

Answer 1

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

The correct options are
A centre is $(1,2)$
B eccentricity is $\frac{4}{5}$
C end points of latus-rectum are $(5,\frac{19}{5})\text{and}(5,\frac{1}{5})$
D length of major axis will be $10$ units

$9x^2+25y^2-18x-100y-116=0$
$\Rightarrow 9(x^2-2x)+25(y^2-4y)=116$
$\Rightarrow 9(x^2-2x+1)+25(y^2-4y+4)=116+100+9$
$\Rightarrow 9(x-1{)}^2+25(y-2{)}^2=225$
$\Rightarrow \frac{(x-1{)}^2}{25}+\frac{(y-2{)}^2}{9}=1$
Hence centre will be $(1,2)$
Comparing it with $\frac{X^2}{a^2}+\frac{Y^2}{b^2}=1$, we get $a=5$ and $b=3$
Here $a>b,$ so the major and the minor axes of the ellipse are parallel to $x-$axis and $y-$axis, respectively.
$\therefore e=\sqrt{1-\frac{b^2}{a^2}}$
$\Rightarrow e=\sqrt{1-\frac{9}{25}}$
$\Rightarrow e=\frac{4}{5}$
end points of latus-rectum will be
$(1+ae,2\pm \frac{b^2}{a})\equiv (5,\frac{19}{5}),(5,\frac{1}{5})$