Write the equation of the ellipse?

Question

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

Answer 1

Your equation is for the 'circle'.
The general equation for an ellipse is:-

x^2/a^2 + y^2/b^2 = 1

Where 'a' is half the major axis, and 'b' is derived from the squares of half the major axis and the square of the distance along the major axis between the centre of the ellipse and one of the foci. (b^2 = a^2 - c^2).

Answer 2

This is an equation of an ellipse. I assume you want it in standard form.

Complete the squares for each variable.

4x² + y² + 16x + 6y - 3 = 0
(4x² + 16x) + (y² + 6y) = 3
4(x² + 4x) + (y² + 6y) = 3
4(x² + 4x + 4) + (y² + 6y + 9) = 3 + 4*4 + 9
4(x + 2)² + (y + 3)² = 28

Divide by 28 to put in standard form.

(x + 2)²/7 + (y + 3)²/28 = 1

Answer 3

Do you mean write this in the standard form of the equation of an ellipse? (That would be (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h,k) is the center, one semi-axis is a and the other is b.) If so, use completion of squares:

4(x^2 + 4x) + (y^2 + 6y) = 3
4(x^2 + 4x + 4) + (y^2 + 6y +9) = 3 + 4(4) + 9 = 28
4(x + 2)^2 + (y + 3)^2 = 28. Now divide both sides by 28 and write denominators as squares:
(x + 2)^2/(sqrt(7))^2 = (y + 3)^2/(2sqrt(7))^2 = 1.

Answer 4

To get in standard form:

Group x's together and y's together:

(4x^2+16x+____)+(y^2+6y+_____)=3

Pull out 4:

4(x^2+4x+____)+(y^2+6y+_____)=3

Add in to get perfect squares (half of x term and ^2 that #) & balance on the other side of the = sign (remember to distribute out the 4 to balance)

4(x^2+4x+_4_)+(y^2+6y+_9__)=3 + 16 + 9


Factor:

4(x+2)^2+(y+3)^2=28

Divide by the 28 to get a 1 on the right hand side of = sign:

4(x+2)^2/28+(y+3)^2/28=28/28

(x+2)^2/7+(y+3)^2/28=1

This is hard because it is hard to see in the format I had to type in in.

Hope this helps.