Given standard equation of ellipse, x 2 a 2 + y 2 b 2 = 1 , a > b , with eccentricity e Match the following a)Focus i) (ae,0) b)Directrix ii) (a,0) c)Eccentricity iii) x= a e d)Vertices iv) (-ae,0) v) x= -a e vi) √ 1- b 2 a 2 Options: A . &nbsp a = i, iv, b = ii, v ,c = vi, d =iii B . &nbsp a = i, b = ii, v ,c = vi, d =iii C . &nbsp a = i, iv, b = iii, v ,c = vi, d = ii D . &nbsp a = i, b = iii, v ,c = vi, d = ii

Answer: Option C : C Let's deduce what all we can infer from ⇒ x 2 a 2 + y 2 b 2 = 1 → f o c u s = ( a e , 0 ) , ( − a e , 0 ) → d i r e c t r i x ⇒ x = a e , x = − a e → e c c e n t r i c i t y ⇒ √ 1 − b 2 a 2 → m a j o r A x i s ⇒ y = 0 a s a > b → m i n o r A x i s ⇒ x = 0 a s b < a → V e r t i c e s ⇒ ( a , 0 ) , ( − a , 0 ) T h e r e f o r e , a → i , i v b → i i i , v c → v i d → i i

Given standard equation of ellipse, x2a2+y2b2=1,a>b, with eccentricity e M

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

Options:

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: Option C
:
C

Let's deduce what all we can infer from $\Rightarrow \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$\to f o c u s = (a e, 0), (- a e, 0) \to d i r e c t r i x \Rightarrow x = \frac{a}{e}, x = \frac{- a}{e} \to e c c e n t r i c i t y \Rightarrow \sqrt{1 - \frac{b^{2}}{a^{2}}} \to m a j o r A x i s \Rightarrow y = 0 a s a > b \to m i n o r A x i s \Rightarrow x = 0 a s b < a \to V e r t i c e s \Rightarrow (a, 0), (- a, 0) T h e r e f o r e, a \to i, i v b \to i i i, v c \to v i d \to i i$