Question

If the latusrectum of an ellipse is equal to half of minor axis, then find it's eccentricity.

Answer 1

Hi

Here is your answer,

Consider the equation of the ellipse is X²/a² + y²/b² = 1

Length of major axis = 2a
Length of minor axis = 2b

and length of latusrectum = 2b²/a

Given that,    

2b²/a = 2b/2

→ a=2b → b = a/2

⇒ b² = a² ( 1-e²)

⇒ [a/2] = a² (1-e²)

⇒ a²/4 = a² ( 1-e²) 

⇒ 1-e² = 1/4

⇒ e²= 1-1/4

⇒ eccentricity = √3/4 = √3/2

Hope it helps you !

Answer 2

AnswEr:

Let the equation of the ellipse be

$\\ \qquad \sf \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1$

and let e be its eccentricity.

It is given that :

$\qquad \sf Latusrectum = \frac{1}{2} \quad(minor \: axis) \\ \\ \\ \implies \sf \: \frac{2 {b}^{2} }{a} = \frac{1}{2} \: (2b) \\ \\ \\ \implies \sf \: 2b = a \\ \\ \\ \implies \sf \: 4 {b}^{2} = {a}^{2} \\ \\ \\ \implies \sf \: 4 {a}^{2} (1 - {e}^{2} ) = {a}^{2} \\ \\ \\ \implies \sf \: 4 - 4 {e}^{2} = 1 \\ \\ \\ \implies \sf \: 4 {e}^{ 3} = 3 \\ \\ \\ \implies \sf \blue {e = \frac{ \sqrt{3} }{2} }$