math homework help?

Question

name type of conic equation and write in standard form

1. x^2+14x-4y+1=0
2. x^2+y^2-14x+16y-4=0
3.4x^2-3y^2+8x+16=0
4. x^2+4y^2+6x-8y+9=0

solve this

Answer 1

Asuming all variables are on the same side of the = sign, if only one variable is squared, it's a parabola. If both are squared and the coefficient of x^2 = the coefficient of y^2, it's a circle. If both are squared but the coefficient of x^2 has the opposite sign of that of y^2, it's a hyperbola. Finally, if both are squared ahd their coefficients differ but have the same sign, it's an ellipse.

So parabola, circle, hyperbola, ellipse

If by standard form you mean graphing form, you have to complete the squares for all squared variables. For example, on #1, change it to x^2 + 14x = 4y - 1

(x+7)^2 = 4y - 1 + 49 = 4y + 48 = 4(y+12)

(x+7)^2 = 4(y+12)

like that

Answer 2

1. parabola
x² + 14x + 49 = 4y - 1 + 49
(x + 7)² = 4(y + 12)

2. circle
x² - 14x + 49 + y² + 16y + 64 = 4 + 49 + 64
(x - 7)² + (y + 8)² = 117

3. hyperbola
4x² + 8x - 3y² = -16
4(x² + 2x + 1) - 3y = -16 + 4
4(x + 1)² - 3y² = -12
y² / 4 - (x + 1)² / 3 = 1

4. ellipse
x² + 6x + 9 + 4y² - 8y = -9 + 9
(x + 3)² + 4(y² - 2y + 1) = 0 + 4
(x + 3)² + 4(y - 1)² = 4
(x + 3)² / 4 + (y - 1)² / 1 = 1