Question

Given standard equation of ellipse,
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b$,
with eccentricity $e$
Match the following
$\begin{array}{cc}\text{a) Focus}& i\left)\right(ae,0)\\ \text{b) Directrix}& ii\left)\right(a,0)\\ \text{c) Eccentricity}& iii)x=\frac{a}{e}\\ \text{d) Vertices}& iv\left)\right(-ae,0)\\ & v)x=\frac{-a}{e}\\ & vi)\sqrt{1-\frac{b^2}{a^2}}\end{array}$

• A. $a=i,iv,b=ii,v,c=vi,d=iii$
• B. $a=i,b=ii,v,c=vi,d=iii$
• C. $a=i,iv,b=iii,v,c=vi,d=ii$
• D. $a=i,b=iii,v,c=vi,d=ii$

Answer 1

The correct option is C

$a=i,iv,b=iii,v,c=vi,d=ii$

Let's deduce what all we can infer from the given equation of the ellipse
$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\to \text{focus}=(ae,0),(-ae,0)\to \text{directrix}\Rightarrow x=\frac{a}{e},x=\frac{-a}{e}\to \text{eccentricity}\Rightarrow \sqrt{1-\frac{b^2}{a^2}}\to \text{major Axis}\Rightarrow y=0asa>b\to \text{minor Axis}\Rightarrow x=0asb<a\to \text{Vertices}\Rightarrow (a,0),(-a,0)\therefore a\to i,iv,b\to iii,v,c\to vi,d\to ii$