What is this equation circle or ellipse?

Question

4x^2+3y^2-16x+30y+79=0


A and B are not equal but x^2 and y^2 is. so it's a ellipse. but it should be a CIRCLE. how? explain

Answer 1

To be a circle, it has to be of the form (x-x_0)^2 + (y-y_0)^2 = r^2.

To be an ellipse, it has to be of the form (x-x_0)^2/r_1^2 + (y-y_0)^2/r_2^2 = 1

(4x^2 - 16x) + (3y^2 + 30y) = -79
4(x^2 - 4x) + 3(y^2 + 10y) = -79
complete the square for each term.

4(x^2 - 4x + 4) + 3(y^2 +10 y + 25) = -79 + 16 + 75 = 12
4(x-2)^2 + 3(y+5)^2 = 12
(x-2)^2/3 + (y+5)^2/4 = 1

So, it's an ellipse with center at (2, -5), x-axis = sqrt(3) y-axis = 2.

Answer 2

4x^2+3y^2-16x+30y+79=0
4(x^2 -4x) +3(y^2+10y) +79 = 0
4(x^2-4x +4) +3(y^2 +10y +25) +79 = 16+75
4(x-2)^2 +3(y+5)^2 =12
(x-2)^2/3 +(y+5)^2/4 =1
This is the equation of an ellipse.
A=4 and B=3 and they are not equal as you state. But your assertion x^2 = y^2 is false and of no significance.

Why do you think it should be a circle? There is no evidence to suggest this. It is definetly an ellipse.

Answer 3

a million) 4x^2 + 9y^2 - 16x +18y -11 = 0 2) 4x^2 - 9y^2 - 16x +18y -11 = 0 3) x^2 + y^2 - 16x + 18y - 11 = 0 once you haven't any longer have been given any time era xy that is plenty extra convenient you seem only the two squares a million) 4x^2 + 9y^2 = 0 has no actual strategies --> ellipse, because of the fact coefficient are no longer equivalent like in 3) x^2 + y^2 - 16x + 18y - 11 = 0 that's the equation of a circle N.B: 4x^2 + 4y^2 = 25 is a circle, too equation 2 represents an hyperbola because of the fact 4x^2 - 9y^2 = 0 has actual strategies (2x + 3y)(2x-3y)=0 2x+3y=0 2x-3y=0 are 2 lines passing interior the process the beginning and parallel to the two asymptotes of the hyperbola you only would desire to locate the middle and you're completed Parabola would nicely be put in the kind (ax + by using)^2 + cx + dy +e = 0 ax+by using=0 is a line parallel to the symmetry axis of the parabola, and with some trouble-free formula would nicely be circled and translated to get the well-known kind did you already know that 2d degree equations in x and y would nicely be degenerate conics like 9x^2 - 16y^2 = 0 that's a pair of lines 3x + 4y = 0 3x - 4y =0 or (x - 3)^2 + (y - 5)^2 = 0 that's a element, specifically the element (3,5) that isn't any longer unusual conic section are referred to as in this way because of the fact they're curves intersection of a (countless) cone and a airplane if the airplane passes interior the process the cone vertex, you get the freak conics i become conversing approximately desire that is useful

Answer 4

first start by writing the equation in standard form instead of general form, it is easier to see what we are working with that way. general form looks like:

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

if the sign between the two terms is positive, its an ellipse, if its negative, its a hyperbola.... if a and b are the same, it is a circle, if a and b are different, its an ellipse. If the coefficient under the x term is bigger than whats underthe y term, then the ellipse has a greater radius on the x axis (stretched horizontally) and the reverse is true for y radius (vertical stretch)

Answer 5

It is an ellipse because the coefficient of x^2 is 4, and the coefficient of y^2 is 3. They are different. Therefore, it cannot be a circle.

Answer 6

4x^2+3y^2-16x+30y+79=0
4x^2 -16x + 3y^2+30y+79=0
(2x-4)^2+(root(3)y+root(3)5)^2 -146 =0

(2x-4)^2+(root(3)y+root(3)5)^2 =146

so it's an elipse

Answer 7

if AC>0 then it's an ellipse. So it's not a circle. If A=C, but does not equal 0, it;s a circle.

Answer 8

I haven't got anyy graph paper to figure this out



....sorry