Question

How do you find center, vertex, and foci of an ellipse `x^2 + 4y^2 = 1`?

Answer 1

`C: (0, 0)`
`V: (+- 1, 0)`
`f: (+-sqrt3/2, 0)`

Explanation:

The standard equation of an ellipse is either in the form

`(x - h)^2/a^2 + (y - k)^2/b^2 = 1`

or

`(x - h)^2/b^2 + (y - k)^2/a^2 = 1`

where `a > b`

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In the given equation

`x^2 + 4y^2 = 1`

This is equivalent to

`(x - 0)^2/1^2 + (y - 0)^2/(1/2)^2 = 1`

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Our center is at `(h, k)`

`C: (0, 0)`

Since `a` is under `x`, the major axis is horizontal. The vertex is at

`V: (h +- a, k)`

`V_1: (0 + 1, 0) => (1, 0)`

`V_2: (0 - 1, 0) => (-1, 0)`

Meanwhile, the foci are `c` units from the center.
Where `c^2 = a^2 - b^2`

`=> c^2 = 1^2 - (1/2)^2`

`=> c^2 = 1 - 1/4`

`=> c^2 = (4 - 1)/4`

`=> c^2 = 3/4`

`=> c = sqrt3/2`

`f: (h +- c, k)`

`f_1: (0 + sqrt3/2, 0) => (sqrt3/2, 0)`

`f_2: (0 - sqrt3/2, 0) => (-sqrt3/2, 0)`