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

• 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 $\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\to focus=(ae,0),(-ae,0)\to directrix\to x=\frac{a}{e},x=\frac{-a}{e}\to eccentricity\Rightarrow \sqrt{1-\frac{b^2}{a^2}}\to majorAxis\Rightarrow y=0asa>b\to minorAxis\Rightarrow x=0asb<a\to Vertices\Rightarrow (a,0),(-a,0)\to Doubleordinate\Rightarrow \text{perpendicular chord to major Axis}\Rightarrow x=0\left(\text{amongst given options passing through ellipse}\right)\text{Latus rectum is defined as perpendicular chord through focus}Weknow,\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1Focus=(ae,0)$
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
$\Rightarrow \frac{(ae{)}^2}{a^2}+\frac{y^2}{b^2}=1y^2=b^2(1-e^2)=a^2(1-e^2)(1-e^2)y=a(1-e^2)\therefore \text{length of latus Rectum}=2a(1-e^2)a=iib=iiic=iiid=i$