Question

Finding vertices for x^2/7 + y^2/5 = 1, the original equation was 5x^2+7y^2=35

Answer 1

you got the center at \((0,0)\) right?

Answer 2

Uhm I think? I'm just supposed to get the vertices before I graph the ellipse

Answer 3

ok lets back up a second
if you want to find the vertices you have to know what the center is first

Answer 4

since this has just \(x^2\) and \(y^2\) and not for example \((x-2)^2\) or something similar, the center is at \((0,0)\)

Answer 5

Ohh okay now I get what you mean by the center

Answer 6

once you have
\[\frac{x^2}{7}+\frac{y^2}{5}=1\] you know that since \(7>5\) the ellipse looks something like this

Answer 7

so..the vertices itself..is it like (+/- 7,0) and (+/-0,5)?

Answer 8

now that we know it looks like the one on the left, we can find the vertices easily
\[\frac{x^2}{7}+\frac{y^2}{5}=1\] is in the form
\[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\] where \(a=\sqrt 7\)

Answer 9

your first answer is right, there are only two vertices not 4
that is why you need to know how it looks first

Answer 10

they are \((-\sqrt7,0)\) and \((\sqrt7,0)\)

Answer 11

got it?

Answer 12

I don't get why they are \[\sqrt{7}\] instead of just 7

Answer 13

the form is
\[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\]

Answer 14

if you want to know "why" for real, you have to derive the formula from the definition, but if you want to know "why" as in "why is the answer \(\sqrt7\) it is because \(a^2=7\) and so \(a=\sqrt7\)

Answer 15

ohh okay

Answer 16

So the 5 doesn't matter then? Whenever he graphed them on the board he used 4 vertices so we had an easier time of drawing

Answer 17

you can also use the \(\sqrt{5}\) i guess

Answer 18

yeah in fact you do need them to get the correct picture

Answer 19

it is just usually the "vertices" means the two points at the longest axis