3x^2+4y^2-6x+16y-15=0 find the elipse

Question

How to find the elipse with the equation in general form

Answer 1

Given equation is 3x^2+4y^2-6x+16y-15 = 0

3x^2-6x+4y^2+16y-15 = 0

3x^2-6x+4y^2+16y-36+19 = 0

3x^2-6x+4y^2+16y+16+3-36 = 0

3x^2-6x+3+4y^2+16y+16-36 = 0

3(x^2-2x+1)+4(y^2+4y+4)-36 = 0

3(x-1)^2+4(y+2)^2-36 = 0

Divide to each side by 36.

3(x-1)^2/36+4(y+2)^2/36 -36/36 = 0/36

(x-1)^2/12+(y+2)^2/9-1 = 0

Add 1 to each side.

(x-1)^2/12+(y+2)^2/9-1+1 = 0+1

(x-1)^2/√(12)^2+(y+2)^2/3^2 = 1

(x-1)^2/3.46^2+(y+2)^2/3^2 = 1

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

Where a >b.

General form of given ellipse equation is (x-1)^2/(3.46)^2+(y+2)^2/3^2 = 1.

Comments

Standard form of  is

.

Answer 2

General form of conics is ax^2 + by^2 +cx +dy + e =0.

The ellipse equation

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.

The standard form for an ellipse is in a form = 1, So divide both sides of equation by 34 to set it equal to 1.

Compare it to standard form of ellipse

a ² > b ²

If the larger denominator is under the "x " term, then the ellipse is horizontal.

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

a  = length of semi-major axis

b  = length of semi-minor axis

Ellipse equation in standard form .