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
$\begin{array}{cc}a)\text{Major axis}& i\left)2a\right(1-e^2)\\ b)\text{Minor axis}& ii)y=0\\ c)\text{Double ordinate}& iii)x=0\\ d)\text{Latus Rectum length}& iv)x=\frac{-a}{e}\\ & v)\sqrt{1-\frac{b^2}{a^2}}\end{array}$

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

Answer 1

The correct option is A

$a=ii,b=iii,c=iii,d=i$

Let's deduce what all we can infer from
$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\to \text{focus}=(ae,0),(-ae,0)\to \text{directrix}\to x=\frac{a}{e},x=\frac{-a}{e}\to \text{eccentricity}\Rightarrow \sqrt{1-\frac{b^2}{a^2}}\to \text{major Axis}\Rightarrow y=0asa>b\to \text{minor Axis}\Rightarrow x=0asb<a\to \text{Vertices}\Rightarrow (a,0),(-a,0)\to \text{Double ordinate}\Rightarrow \text{perpendicular chord to major Axis}\Rightarrow x=0\left(\text{amongst given options passing through ellipse}\right)\text{Latus rectum is defined as perpendicular chord through focus}\text{We know},\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{Focus}=(ae,0)$
Let's calculate the $y$ coordinates of this chord. We know $x$ coordinate is (ae,o) because it's the focus.
Substituting in ellipse equation
$\Rightarrow \frac{(ae{)}^2}{a^2}+\frac{y^2}{b^2}=1y^2=b^2(1-e^2)=a^2(1-e^2)(1-e^2)y=a(1-e^2)\therefore \text{length of latus Rectum}=2a(1-e^2)a=iib=iiic=iiid=i$