How do I graph $\frac{{(x + 2)}^{2}}{9} + \frac{{(y - 3)}^{2}}{25} = 1$ $(x+2)^2/9+(y-3)^2/25=1$ algebraically?

Question

How do I graph $\frac{{(x + 2)}^{2}}{9} + \frac{{(y - 3)}^{2}}{25} = 1$ $(x+2)^2/9+(y-3)^2/25=1$ algebraically?

1 Answer

Impact of this question

3269 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-10-08T16:47:10

You have an ellipse in the form

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

where (h, k) is the ellipse's center.
You have $(- 2, 3)$ $(-2, 3)$ as your center.

$a$ $a$ is under $y$ $y$ . That means your major axis is vertical with length $2 a = 10$ $2a = 10$ .
The endpoints of your major axis are $(- 2, - 2)$ $(-2, -2)$ and $(- 2, 8)$ $(-2, 8)$

Meanwhile, the minor axis is horizontal with length $2 b = 6$ $2b = 6$
The endpoints of your major axis are $(- 5, 3)$ $(-5, 3)$ and $(1, 3)$ $(1, 3)$

Connect the endpoints of your major and minor axis and you have your ellipse

Science

Math

Humanities

... and beyond

How do I graph $\frac{{(x + 2)}^{2}}{9} + \frac{{(y - 3)}^{2}}{25} = 1$ $(x+2)^2/9+(y-3)^2/25=1$ algebraically?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do I graph (x+2)29+(y−3)225=1(x+2)^2/9+(y-3)^2/25=1 algebraically?

1 Answer

Related questions

Impact of this question

How do I graph $\frac{{(x + 2)}^{2}}{9} + \frac{{(y - 3)}^{2}}{25} = 1$ $(x+2)^2/9+(y-3)^2/25=1$ algebraically?