Conics Algebra problem?
"Determine values for A, B, and C such that the equation below represents the conic with a horizontal axis and a vertical axis. Then rewrite your equation for each conic in standard form, identify (h,k) and describe the translation:
Ax^2+Bxy+Cy^2+2x-4y+5=0
a. Circle
b. Ellipse
c. Parabola
d. Hyperbola"
If anyone could at least help give me an idea of exactly what I'm supposed to do, it would be greatly appreciated.
Comments
Ax² + Bxy + Cy² + 2x - 4y + 5 = 0
First let B = 0, otherwise the axes of the conics won't be parallel to the x- and y-axes.
c. A parabola has exactly one squared variable, so either A = 0 or C = 0 (but not both).
Let C = 0.
rearrange to solve for y
y = (A/4)x² + (1/2)x + 5/4
factor out leading coefficient
y = (A/4)(x² + (2/A)x) + 5/4
Might as well make it simple and let A = 4.
y = (x² + (1/2)x) + 5/4
complete the square
y = (x² + (1/2)x + (1/4)²) - (1/4)² + 5/4
= (x + 1/2)² + 19/16
This is an up-opening parabola with vertex (-1/2, 19/16).
you will could desire to end the sq. one after the different for the x and y factors, so first rearrange to place the x's and y's mutually. 16x^2 + 64x + 9y^2 - 18y - seventy one = 0 sixteen(x^2 + 4x) + 9(y^2 - 2y) - seventy one = 0 sixteen(x^2 + 4x + 4 - 4) + 9(y^2 - 2y + a million - a million) - seventy one = 0 sixteen(x^2 + 4x + 4) - sixty 4 + 9(y^2 - 2y + a million) - 9 - seventy one = 0 sixteen(x+2)^2 + 9(y+a million)^2 - a hundred and forty four = 0 This equation is commencing as much as look like an ellipse. sixteen(x+2)^2 + 9(y+a million)^2 = a hundred and forty four (x+2)^2 / 9 + (y+a million)^2 / sixteen = a million (x+2)^2 / 3^2 + (y+a million)^2 / 4^2 = a million The equation is an ellipse with a considerable axis of four contraptions vertically, and a minor axis of three contraptions horizontally, based at (-2,-a million).