7x^2+2y^2+14x-8y+1=0 write equation in standard form for the conics described

Question

this is the instruction given: Reduce the equation  of the conics to the standard form and draw the curve

Answer 1

Given equation 7x^2+2y^2+14x-8y+1 = 0

add 14 to each side.

7x^2+14x+2y^2-8y+1+14 = 0+14

7x^2+14x+2y^2-8y+7+8 = 14

7x^2+14x+7+2y^2-8y+8 = 14

7(x^2+2x+1)+2(y^2-4y+4) =14

7(x+1)^2+2(y-2)^2 = 14

Divide to each side by 14.

7(x+1)^2/14+2(y-2)^2/14 = 14/14

(x+1)^2/2+(y-2)^2/7 = 1

(x-(-1))^2/2+(y-2)^2/7 = 1

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

 

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.

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 (0.41, 2)

Left most point (-2.41, 2)

Top most point (-1,4.64)

Bottom most point (-1, -0.64)

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