Find the center, vertices and foci for the ellipse 4x^2 + 16y^2 = 64

3 Answers | Add Yours

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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)

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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).

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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).

We’ve answered 318,928 questions. We can answer yours, too.

Ask a question