how do i find the foci of an elipse?

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.

A community for students.

See more answers at brainly.com
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 this expert

answer on brainly

SEE EXPERT ANSWER

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

do you know where the center and vertexes are?
if so, then the foci are an odd construction of the usual pythag thrm ... in this case, b and c are legs, and a is a hypotenus giving us a rather strange look of c^2 + b^2 = a^2, such that x is the distance from center to focus
well the equation i have is (x+2)^2/9+(y-4)^2/36=1

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

then we know a^2 and b^2, right?
yes
so show me c^2 :)
36+ =9 c^2=25
-25
36 - 9 = 27
oh yeah lol
so 3cbrt3 = c, or this distance from center.
3 sqrt3 lol
so, tell me our center, and in what direction we need to move
center (-2,4) we move left and right
not left and right notice that under y is bigger than under x y is bigger than x, so our focuses are going to be in relation to y |dw:1433524715502:dw|
so (-2, 4+ or - 3sqrt3)
yes
alright
thanks:)
youre welcome
can you help me with one other thing
depends on what it is and how good my memory is :)
finding the eccentricity of (x+7)^/16+(y-3)^2/4=1
define the formula for eccentricity
eccentricity = c/a
thats what i thought so sqrt(c^2/a^2) should work .. c^2 is the difference of the bottoms, and a^2 is the larger of them
So sqrt 12/4
um, 16 is bigger than 4
because sqrt 16 is 4
oh, youve got some notation off then sqrt(12/16) sqrt(3/4) sqrt(3)/2
Wait I dont get how you reduced those? wouldnt it just be sqrt(3/4)
Like why did you reduce so far?
thats what you do with fractions, you reduce them till they have no common factors
makes sense.
So final answer is sqrt (3/2)
\[\sqrt{\frac{16-4}{16}}\] \[\sqrt{\frac{12}{16}}\] \[\sqrt{\frac{4*3}{4*4}}\] \[\sqrt{\frac{3}{4}}\] \[\frac{\sqrt3}{\sqrt4}\]
then reduce? or
can it be simplified more?
i just showed you some of the process, i didnt finish it ... i figured youd be cognizant enough to see where its going.
yea sqr t4 is equal to 2
then lets go with sqrt(3) /2
Then you have sqrt(3/2) :) thanks
no, not sqrt (3/2) im not sure if its a typing error of yours, or if you are making a mistake. it is: sqrt(3) ------ 2
sqrt(3) sqrt(3) ----- = ----- sqrt(4) 2
yeah thats how I percieved it and meant it
good luck :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question