Question

find the eccentricity,coordinates of foci,length of the latus rectum of the ellipse 4x^2+9y^2= 1

Answer 1

eccentricity=√5/3
foci
(1/2,0)
(-1/2,0)

Answer 2

Eccentricity = √5/3, co-ordinates of foci = (√5/6, 0) and (-√5/6, 0) and length of the latus rectum of the ellipse = 4/9.

Step-by-step explanation:

Given:

The given equation of ellipse is $4x^2 + 9y^2 = 1$

Comparing this equation with equation of ellipse in standard form,

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

We get,

$a^2=\frac{1}{4}$ ⇒ $a=\sqrt{\frac{1}{4} }$⇒$a={\frac{1}{2} }$  

and  $b^2=\frac{1}{9}$  ⇒ $a=\sqrt{\frac{1}{9} }$  ⇒$a={\frac{1}{3}}$  

∴ Eccentricity  (e)

= $\sqrt{1-(\frac{b}{a})^2 }$

= $\sqrt{1-\frac{(\frac{1}{3})^2}{(\frac{1}{2})^2 } }$

= $\sqrt{1-(\frac{2}{3})^2 }$

= $\sqrt{1-\frac{4}{9} }$

= $\sqrt{\frac{9-4}{9} }$

= $\sqrt{\frac{5}{9} }$

= $\frac{\sqrt{5} }{3}$

Coordinates of foci are (ae, 0) and (– ae, 0)

= $(\frac{1}{2}\times\frac{\sqrt{5} }{3}, 0) and (-\frac{1}{2}\times\frac{\sqrt{5} }{3}, 0)$

= $(\frac{\sqrt{5} }{6}, 0 ) and (-\frac{\sqrt{5} }{6}, 0 )$

And length of of latus rectum

= $\frac{2b^2}{a}$

= $(\frac{2\times(\frac{1}{3})^2 }{\frac{1}{2} } )$

= $2\times2\times\frac{1}{3}\times \frac{1}{3}$

= $\frac{4}{9}$