what is the foci, center, and vertices of the ellipse?

Question

4x^2+y^2-32x+4y+64=0

Answer 1

Given eqution is 4x^2+y^2-32x+4y+64 = 0

4x^2-32x+64+y^2+4y = 0

Add 4 to each side.

4x^2-32x+64+y^2+4y+4 = 0+4

4(x^2-8x+16)+y^2+4y+4 = 4

4(x-4)^2+(y+2)^2 = 4

Divide to each side by 4.

4(x-4)^2/4+(y+2)^2/4 = 4/4

(x-4)^2/1^2+(y+2)^2/2^2 = 1

(x-4)^2/1^2+(y-(-2))/2^2 = 1

Compare it to standard form of ellpse equation.

(x-h)^2/b^2+(y-k)^2/a^2 = 1

Where a > b.

a = 2, b = 1.

Center of ellipse is (h,k) = (4,-2)

c = √(a^2-b^2) = √4-1

c = √3

Vertices are (h-b,k), (h+b,k)

(4-1,-2),(4+1,-2)

Vertices are (3,-2)(5,-2).

Foci (h,k+c)(h,k-c)

Foci(4,-2+√3)(4,-2-√3)

Comments

Vertices of 4x ²+ y ²- 32x + 4y + 64 = 0  are (4,0),(4,-4).

Answer 2

The equation is

To change the expressions (x ²- 8x ) and (y ² + 4y ) into a perfect square trinomial,

add (half the x coefficient)² and add (half the y coefficient)²to each side of the equation.

The standard form for an ellipse is in a form = 1, So divide both sides of equation by 4 to set it equal to 1.

Compare the standardform of ellipse

a ² > b ²

If the larger denominator is under the "y " term, then the ellipse is vertical.Center (h, k ).

a  = length of semi-major axis, b  = length of semi-minor axis.

vertices: (h , k + a ), (h, k - a )

co-vertices: (h + b , k ), (h - b, k ) [endpoints of the minor axis]

"c" is the distance from the center to each focus.

Foci: (h , k + c ), (h , k - c )

a  = 2, b  = 1

(h , k ) = (4, -2)

Vertices: (4, -2 + 2), (4, -2 - 2)

(4,0),(4,-4)

Center (4, -2)

Vertices (4,0),(4,-4)

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