find the vertices and the foci of (x-5)^2/4+(y+3)/16=1.

Question

samsan9 · Answer

$$\frac{ (x-5)^2 }{ 4 }+\frac{ (y+3)^2 }{16  }=1$$

samsan9 · Answer

this?

anonymous · Answer

Yup one sec. Let me solve it.

anonymous · Answer

So the first thing you have to do is find the center right? Which is (h,k) so what is the center?

samsan9 · Answer

(5,-3)

anonymous · Answer

Ok so is it a vertical elipse or horizontal?

samsan9 · Answer

vertical i think?

anonymous · Answer

Yup so in order to find how high it goes above the center you need to find the square root of 16 which is 4. So for the vertices they will be 4 above and 4 below the center point right?

samsan9 · Answer

yes i think so

anonymous · Answer

Ok so what are your vertices then?

samsan9 · Answer

(2,4)

anonymous · Answer

No you have two sets of vertices and so one will be 4 above (5,-3) and one will be below (5,-3) so they would be (5,1) and (5,-7)

anonymous · Answer

So now you have to find the foci which is represented by the equation c^2=a^2-b^2 and since 16 is A because it's the bigger number and 4 is b because it is the smaller number you just plug them into the equation.

anonymous · Answer

So what is 16-4?

samsan9 · Answer

12

anonymous · Answer

Ok and since it's c squared you have to find the square root of 12 but since there is no perfect root c=plus or minus the square root of 12. And that is your foci. So just remember the equation c^2=a^2-b^2 and also that in the equation the bigger number is always a^2 and the smaller number is always b^2.

anonymous · Answer

So Vertices=(5,1) and (5,-7)
Foci=

anonymous · Answer

plus or minus sqrt of 12

samsan9 · Answer

do you think i could tag you in another problem of mine?

anonymous · Answer

Sure i'll try to help.

samsan9 · Answer

it might be a little long... but it is one of my last problems on here

anonymous · Answer

Ok