Question

$If{e}_{1}and{e}_{2}arerespectivelytheeccentricitiesoftheellipse\frac{x^2}{18}+\frac{y^2}{4}=1andthehyperbola\frac{x^2}{9}-\frac{y^2}{4}=1,thentherelationbetween{e}_{1}and{e}_{2}is$

• A. $3e_1^2+e_2^2=2$
• B. $e_1^2+2e_e^2=3$
• C. $2e_1^2+e_2^2=3$
• D. $e_1^2+3e_2^2=2$

Answer 1

The correct option is C

$2e_1^2+e_2^2=3$

$\text{The standard form of the ellipse is}$

$\frac{x^2}{18}+\frac{y^2}{4}=1,where{a}^2=18and{b}^2=4$

$\text{So,the eccentricity is calculated in the following way:}$

$b^2{=}^2(1-e_1^2)$

$\Rightarrow 4=18(1-e_1^2)$

$\Rightarrow \frac{2}{9}=1-2_1^2$

$\Rightarrow e_1^2=\frac{7}{9}$

$\text{The standard form of the hyperbola is}$

$\frac{x^2}{9}+\frac{y^2}{4}=1,where{a}^2=9and{b}^2=4$

$\text{So,the eccentricity is calculated in the following way:}$

$b^2=a^2(e_2^2-1)$

$\Rightarrow 4=9(e_2^2-1)$

$\Rightarrow \frac{4}{9}=e_2^2-1$

$\therefore 2e_1^2+e_2^2=\frac{2\times 7}{9}+\frac{13}{9}$

$=\frac{27}{9}=3$