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
$\begin{array}{cc}\text{a)Focus}& \text{i) (ae,0)}\\ \text{b)Directrix}& \text{ii) (a,0)}\\ \text{c)Eccentricity}& \text{iii) x=}\frac{\text{a}}{\text{e}}\\ \text{d)Vertices}& \text{iv) (-ae,0)}\\ & \text{v) x=}\frac{\text{-a}}{\text{e}}\\ & \text{vi)}\sqrt{\text{1-}\frac{{\text{b}}^{\text{2}}}{{\text{a}}^{\text{2}}}}\end{array}$

• 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 $\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\to focus=(ae,0),(-ae,0)\to directrix\Rightarrow x=\frac{a}{e},x=\frac{-a}{e}\to eccentricity\Rightarrow \sqrt{1-\frac{b^2}{a^2}}\to majorAxis\Rightarrow y=0asa>b\to minorAxis\Rightarrow x=0asb<a\to Vertices\Rightarrow (a,0),(-a,0)Therefore,a\to i,ivb\to iii,vc\to vid\to ii$