Question

Match the following for ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b$, with eccentricity $e.$

• A. $2a(1-e^2)$
• B. $y=0$
• C. $x=0$
• D. $x=2a$

Answer 1

Let's deduce what all we can infer from $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Foci are $(ae,0),(-ae,0)$
Directrices are $x=\pm \frac{a}{e}$
Eccentricity, $e$ is $\sqrt{1-\frac{b^2}{a^2}}$
Vertices are $(a,0),(-a,0)$
Major axis is $y=0$
Minor axis is $x=0$
Double ordinate is a chord perpendicular to the major axis $\Rightarrow x=2a$ is the equation of the double ordinate.
[Please note that $x=0$ also also a double ordinate]
Latus rectum is defined as a chord perpendicular to the major axis and passing through focus.
We know, ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Focus is $(ae,0)$
Let's calculate the $y$ coordinate of this chord. We know $x$ coordinate is $(ae,o)$ because it's the focus.
Substituting in the equation of the ellipse, we get,
$\Rightarrow \frac{(ae{)}^2}{a^2}+\frac{y^2}{b^2}=1$
$\Rightarrow y^2=b^2(1-e^2)=a^2(1-e^2)(1-e^2)$
$\Rightarrow y=a(1-e^2)$
$\therefore$ Length of latus rectum $=2y=2a(1-e^2)$