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

Question

8630 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-20T12:04:06

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

Focii are at $(- 3, 0 and 3, 0)$ $(-3,0 and 3,0)$ So distance of focii from

centre is $c = 3$ $c=3$ , y intercepts are at $(0,-4) and (0,4) :.$

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

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

Relationship among $a, b, and c$ 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 $= 2a =2*5=10$ and minor axis length

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

$x^2/a^2+y^2/b^2=1 or x^2/5^2+y^2/4^2=1$ 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]