CLASSIFYING A CONIC FROM ITS GENERAL EQUATION

Put each of the following equations in standard form and classify the conic.

Problem 1 :

9y² – x² + 2x + 54y + 62 = 0

Solution :

9y² – x² + 2x + 54y + 62 = 0

9y² – x² + 2x + 54y = -62

9y² + 54y – x² + 2x = -62

9(y² + 6y) – (x² - 2x) = -62

9[(y + 3)² - 3²] – [(x - 1)² - 1²] = -62

9[(y + 3)² - 9] – (x - 1)² + 1 = -62

9 (y + 3)² + 81 – (x - 1)² + 1 = -62

9 (y + 3)²  – (x - 1)² + 82 = -62

9 (y + 3)²  – (x - 1)² = -62 - 82

9 (y + 3)²  – (x - 1)² = -144

Dividing by -144, we get 

(x - 1)²/144 - (y + 3)² / 16 = 1

It is in the form of (x - h)²/a² - (y - k)² / b² = 1

It is a hyperbola.

Problem 2 :

4x² + y² - 8x + 4y - 16 = 0

Solution :

4x² + y² - 8x + 4y - 16 = 0

4x² - 8x + y² + 4y = 16

4(x² - 2x) + (y² + 4y) = 16

4 [(x - 1)² + 1] + [(y + 2)² - 2²] = 16

4 [(x - 1)² + 1] + [(y + 2)² - 4] = 16

4 (x - 1)² + 4 + (y + 2)² - 4 = 16

4 (x - 1)² + (y + 2)² = 16

Dividing by 16, we get

(x - 1)²/4 + (y + 2)²/16 = 1

It is in the form of (x - h)²/a² + (y - k)² / b² = 1

It is a ellipse

Problem 3 :

x² + y² + 6x - 4y + 12 = 0

Solution :

x² + y² + 6x - 4y + 12 = 0

x² + 6x + y² - 4y + 12 = 0

(x + 3)² - 3² + (y - 2)² - 2² + 12 = 0

(x + 3)² - 9 + (y - 2)² - 4 + 12 = 0

(x + 3)² + (y - 2)² - 13 + 12 = 0

(x + 3)² + (y - 2)² - 1 = 0

(x + 3)² + (y - 2)²  = 1

It is in the form of (x - h)² + (y - k)² = r²

It is a circle.

Problem 4 :

x² - 2y + 16x + 28 = 0

Solution :

x² - 2y + 16x + 28 = 0

2y = x² + 16x + 28

2y = (x + 4)² - 4² + 28

2y = (x + 4)² - 16 + 28

2y = (x + 4)²  + 12

y = (1/2) (x + 4)²  + 6

It is in the form, y = 4a (x - h)² + k

So, it is parabola.