16x^2+4y^2-32x+16y-32=0 find the ellipse

Question

graph a ellipse

Answer 1

Given 16x^2+4y^2-32x+16y-32 = 0

16x^2-32x+4y^2+16y-32 = 0

16x^2-32x+4y^2+16y+32-64 = 0

16x^2-32x+4y^2+16y+16+16-64 = 0

16x^2-32x+16+4y^2+16y+16 = 64

16(x^2-2x+1)+4(y^2+4y+4) = 64

16(x-1)^2+4(y+2)^2 = 64

Divide to each side by 64.

16(x-1)^2/64+4(y+2)^2/64 = 64/64

(x-1)^2/4+(y+2)^2/16 = 1

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

Compare it to standard form of (x-h)^2/b^2+(y-k)^2/a^2 = 1

When a > b.

Here center is (h,k) = (1,-2)

a = 4, b = 2

Graph to given ellipse.

 

Answer 2

The equation is

To change the expressions (x ²- 2x) 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.

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.

a  = 4 , b  = 2.

The points for this ellipse are ,

Right most point (h +b , k )

Left most point (h - b , k )

Top most point (h , k + a )

Bottom most point (h , k - a )

Right most point (3, -2)

Left most point (-1, -2)

Top most point (1, 2)

Bottom most point (1, -6)

Graph

1. draw the coordinate plane.

2. Plot the center at (1, -2).

3.Plot 4 points away from the center in the up, down, left and right direction.

4.Sketch the ellipse.