Question

$If{e}_{1}istheeccentricityoftheconic9x^2+4y^2=36and{e}^{2}istheeccentricityoftheconic9x^2-4y^2=36,then$

• A. $e_1^2-e_2^2=2$
• B. $2<e_2^2-e_1^2<3$
• C. $e_2^2-e_1^2=2$
• D. $e_2^2-e_1^2>3$

Answer 1

The correct option is B

$2<e_2^2-e_1^2<3$

$Theconic9x^2-4y^2=36canrewritteninthefollowingway.$

$\frac{9x^2}{36}+\frac{4y^2}{36}=1$

$\Rightarrow \frac{x^2}{4}+\frac{y^2}{9}=1$

$\text{This is the standard equation of an ellipse.}$

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

$\Rightarrow 9=4(1-e_1{)}^2$

$\Rightarrow (e_1{)}^2=\frac{-5}{4}$

$Theconic9x^2-4y^2=36canrewritteninthefollowingway:$

$\frac{9x^2}{36}-\frac{4y^2}{9}=1$

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

$\text{This is standard equation of a hyperbola}$

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

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

$\Rightarrow (e_2{)}^2=\frac{13}{4}$

$\therefore e_2^2=\frac{13}{4}+\frac{5}{4}=2.5$