At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

plug in values for x and y?

yeah.. as a check , one of the points is already on the y axis

for the b value

(0,54) is good right?

yes, and (8,18)

0^2/a^2 so that cancels out and I have:
54^2/b^2 = 1

yeah it crosses the y axis at 54, so that checks out for b right

2916/b^2 = 1
how do I simplify?

it will cross the y axis at + and - b

multiply by b^2, then take square root

should just say 54^2 = b^2 , so b is 54

oh lol, which beans I get +-54?

ok and then I plug that back in with the point 8,18 right?

yep, they really gave you that value, but it checks correct

need to find + and - a, where it crosses x axis

use the other point, and the b value, solve a

8^2/a^2 + 18^2/54^2 = 1
64/a^2 + 324/2916= 1 how do I simplify

I got a = 6 * sqrt(2)
Is that right?

so I get:
x^2/72 + y^2/2916 = 1
for the final equation is that right?

[1/3]^2 = 1/9

is my final equation correct?

x^2/72 + y^2/2916 = 1?

yes that is right for a, but remember it is + and -

so how do I change the final equation?

for the standard form, you want to leave those denominators as a^2 and b^2, not expanded out

\[\frac{ x^2 }{ [6\sqrt{2}]^2 }+\frac{ y^2 }{ 54^2 } = 1\]

ok thanks

welcome