# CLASSIFYING A CONIC FROM ITS GENERAL EQUATION

## Equation of Parabola

The variable which is not having square, then the parabola is symmetric about that particular axis.

## Equation of Ellipse

• If a2 is greater then, the ellipse will be symmetric about x-axis.
• If b2 is greater then, the ellipse will be symmetric about y-axis.

## Equation of Hyperbola

Equation of hyperbola which is symmetric about x-axis.

Equation of hyperbola which is symmetric about y-axis.

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)+ 1] + [(y + 2)2 - 22] = 16

4 [(x - 1)+ 1] + [(y + 2)2 - 4] = 16

4 (x - 1)+ 4 + (y + 2)2 - 4 = 16

4 (x - 1)+ (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)- 22 + 12 = 0

(x + 3)2 - 9 + (y - 2)- 4 + 12 = 0

(x + 3)2 + (y - 2)- 13 + 12 = 0

(x + 3)2 + (y - 2)- 1 = 0

(x + 3)2 + (y - 2) = 1

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

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