# How do you find the standard form given vertices Foci (-3,0) and (3,0) and y intercepts (0,-4) and (0,4)?

Jan 20, 2018

Standard form of the ellipse is ${x}^{2} / 25 + {y}^{2} / 16 = 1$

#### Explanation:

Focii are at $\left(- 3 , 0 \mathmr{and} 3 , 0\right)$ So distance of focii from

centre is $c = 3$ , y intercepts are at $\left(0 , - 4\right) \mathmr{and} \left(0 , 4\right) \therefore$

distance of vertices at minor axis from centre is $b = 4$

Let distance of vertices at major axis from centre is $a$.

Relationship among $a , b , \mathmr{and} c$ is ${c}^{2} = {a}^{2} - {b}^{2}$

:. 3^2= a^2-4^2 or a^2 =3^2+4^2=25 :. a=5 ; a > b

Major axis length $= 2 a = 2 \cdot 5 = 10$ and minor axis length

$= 2 b = 2 \cdot 4 = 8$ . Hence standard form of the ellipse is

${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1 \mathmr{and} {x}^{2} / {5}^{2} + {y}^{2} / {4}^{2} = 1$ or

${x}^{2} / 25 + {y}^{2} / 16 = 1$

graph{x^2/25+y^2/16=1 [-14.24, 14.24, -7.12, 7.12]} [Ans]