Find the center, vertices and foci for the ellipse 4x^2 + 16y^2 = 64
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Given the equation of an ellipse is 4x^2 + 16y^2 = 64.
We need to find the center and foci.
First we need to re-write the equation into the standard form of the ellipse.
==> x^2/a^2 + y^2 / b^2 = 1..............(1)
==> 4x^2 + 16y^2 = 64.
Let us divide by 64 .
==> 4x^2/64 + 16y^2/64 = 1
==> x^2/ (64/4) + y^2/ (64/16) = 1
==> x^2 / (16) + y^2 / 4 = 1
==> x^2/ (4^2) + y^2 / (2^2) = 1............(2)
Comparing equation (1) and (2) we conclude that:
a = +-4 and b=+- 2.
Then the vertices's are:
(4,0) , (-4,0), ( 0,2) and ( 0,-2).
==> Then the center is (0, 0).
==> C^2 = a^2 - b^2
==> C^2 = 16 - 4 = 12
==> c= +-sqrt12 = +-2sqrt3
==> The foci is ( 2sqrt3,0 ) and ( -2sqrt3,0)
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To solve To find the centre and vertices of 4x^2+16y^2 = 64.
We divide both sides by 64:
4x^2/64 +16y^2 /64 = 1.
x^2/(64/4) + y^2/(64/16) = 1.
x^2/16 +y^2/4 = 1.
x^2/4^2+y^2/2^2 = 1 is in the standard form of an ellipse, with centre (0,0) and semi-major axis a = 4 and semi-minor axis b = 2.
So the vertices are at (-4,0) and (4,0), and accentricity e = sqrt(1-b^2/a^2) = sqrt(1-4/16) = (sqrt 3)/2.
Therefore focus S is at (-ae ,0) = {4*(sqrt3)/2 , 0) = (2sqrt3,0). The other focus S' = (-ae , 0) = (-2sqrt3, 0).
First, we'll put the equation of the ellipse in the standard form:
(x/a)^2 + (y/b)^2 = 1
To put the given equation in this form, we'll divide by 64 both sides:
4x^2 + 16y^2 = 64
4x^2/64 + 16y^2/64 = 64/64
We'll simplify and we'll get:
x^2/16 + y^2/4 = 1
We'll identify a^2 = 16 => a = +/-4
b^2 = 4
b = +/-2
The vertices of the ellipse are:
(a,0) and (-a,0)
(0,b) and (0,-b)
We'll substitute a and b and we'll get the vertices of the given ellipse: (4,0) ; (-4,0) ; (0,2) ; (0,-2).
The center of the ellipse is (0,0).
The foci of the ellipse are: (c,0) and (-c,0).
We'll compute the coordinate c:
c^2 = a^2 - b^2
c^2 = 16 - 4
c^2 = 12
c1 = 2sqrt3
c2 = -2sqrt3
The foci are: (2sqrt3,0) and (-2sqrt3,0).
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