Question

The arrangement of the following ellipse in the ascending order of their eccentricities
A. $3x^2+4y^2=12$
B. $9x^2+5y^2-30y=0$
C. $x=4\cos \theta ,y=5\sin \theta$
D. $8x^2+9y^2=72$

• A. A,B,C,D
• B. D,A,B,C
• C. B,C,A,D
• D. D,C,B,A

Answer 1

Answer: (B)

D,A,B,C

$\left(A\right)$

$3x^2+4y^2=12$
$\frac{x^2}{4}+\frac{y^2}{3}=1$
$(e_1{)}^2=\frac{4-3}{4}$
$\therefore e_1=\frac{1}{2}$

$\left(B\right)$

$9x^2+5(y-3{)}^2=45$
$\frac{x^2}{5}+\frac{(y-3)}{9}=1$
$(e_2{)}^2=\frac{9-5}{9}$
$\therefore e_2=\frac{2}{3}$

$\left(c\right)$ $x=4\cos \theta ,y=5\sin \theta$
$a=4,b=5$
$(e_3{)}^2=\frac{25-16}{25}$
$(e_3{)}^2=9/25$

$\therefore e_3=\frac{3}{5}$

$\left(D\right)$

$\frac{x^2}{9}+\frac{y^2}{8}=1$

$a^2=9,b^2=8$
$(e_4{)}^2=\frac{9-8}{9}$

$\therefore e_4=\frac{1}{3}$
$⟹e_2>e_3>e_1>e_4$
$⟹\left(B\right)>\left(C\right)>\left(A\right)>\left(D\right)$

Hence, ascending order of these ellipse is $D,A,B,C$.