Question

Find the equation of ellipse whose foci are (+-4,0) and eccentricity is 1/3

Answer 1

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Here we have,

Foci is on the x axis (Given)

Hence,

It is a horizontal ellipse

Now,

Assume

${\boxed{\sf\:{Equation=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1}}}$

Where a² > b²

Assume,

Foci = (±c , 0)

Hence,

c = 4

Also

${\boxed{\sf\:{e=\dfrac{c}{a}}}}$

${\boxed{\sf\:{a=\dfrac{c}{e}}}}$

${\boxed{\sf\:{\dfrac{4}{1/3}}}}$

= 12

Now,

c² = (a² - b²)

b² = (a² - c²)

b² = (144 - 16)

= 128

Therefore,

= a²

= (12)²

= 144

Also,

b² = 128

Hence,

$\Large{\boxed{\sf\:{Equation=\dfrac{x^2}{144}+\dfrac{y^2}{128}=1}}}$

Answer 2

Given ,

The focii of eclipse is ( ± 4 , 0 ) and eccentricity is 1/3

So ,

c = 4

We know that , the eccentricity is given by

e = c/a

Thus ,

1/3 = c/a

1/3 = 4/a

a = 12

Now , the relationship between c , a and b in eclipse is given by

(a)² = (b)² + (c)²

Thus ,

(12)² = (b)² + (4)²

144 = (b)² + 16

(b)² = 128

∴ a > b

The equation of the eclipse will be

$\boxed{ \tt{ \frac{ {(x)}^{2} }{144} + \frac{ {(y)}^{2} }{128} = 1 }}$