How do you find the coordinates of the center, foci, the length of the major and minor axis given $36 x^{2} + 81 y^{2} = 2916$ $36x^2+81y^2=2916$ ?

Question

2176 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-15T00:50:49

You can obtain your answers by doing steps 1 through 4 in the explanation.

Let divide by 2916 and write the denominators as squares:

$\frac{x^{2}}{9^{2}} + \frac{y^{2}}{6^{2}} = 1$ $x^2/9^2 + y^2/6^2 = 1$

When the denominator of the x term is greater than the denominator of the y term, the standard form is:

$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$ $(x - h)^2/a^2 + (y - k)^2/b^2 = 1$

where:

$(h, k)$ $(h, k)$ is the center point
$2 a$ $2a$ is the length of the major axis
$2 b$ $2b$ is the length of the minor axis
The foci are at $(h + \sqrt{a^{2} - b^{2}}, k)$ $(h + sqrt(a^2 - b^2), k)$ and $(h - \sqrt{a^{2} - b^{2}}, k)$ $(h - sqrt(a^2 - b^2), k)$

Subtract zero from x and y to put the equation in standard form:

$\frac{{(x - 0)}^{2}}{9^{2}} + \frac{{(y - 0)}^{2}}{6^{2}} = 1$ $(x - 0)^2/9^2 + (y - 0)^2/6^2 = 1$

You can do steps 1 through 4 for your answer.

How do you find the coordinates of the center, foci, the length of the major and minor axis given 36x2+81y2=291636x^2+81y^2=2916?