IDENTIFYING THE CONICS FROM THE GENERAL EQUATION OF THE CONIC

Example 1 :

Identify the type of conic section for each of the equations.

(i)   2x² − y² = 7

Solution :

By comparing the given equation with the general form of conic Ax² + Bxy + Cy² + Dx + Ey + F = 0, we get 

A = 2, C = -1 and F = -7 

The values of A and C are not equal and it has opposite signs. Hence it is a hyperbola.

(ii)  3x² + 3y² − 4x + 3y  - 10 = 0

By comparing the given equation with the general form of conic Ax² + Bxy + Cy² + Dx + Ey + F = 0, we get 

3x² + 3y² − 4x + 3y  - 10 = 0

Dividing the entire equation by 3, we get

x² + y² − (4/3)x + y  - (10/3) = 0

x² − (4/3)x + y² + y = 10/3

(x - (4/6))² - (4/6)² + (y + (1/2))² - (1/2)²  =  10/3

(x - (2/3))² + (y + (1/2))²  =  10/3 + (4/9) + (1/4)

(x - (2/3))² + (y + (1/2))²  =  (120 + 16 + 9)/36

(x - (2/3))² + (y + (1/2))²  =  145/36

(x - (2/3))² + (y + (1/2))²  =  (√145/6)²

Hence it is a circle.

(iii)  3x² + 2y² = 14

Solution :

3x² + 2y² = 14

A ≠ C and A and C are of the same sign, the general equation yields an ellipse.

(iv)  x² + y² + x − y = 0

Solution :

x² + x + y²− y = 0

(x + (1/2))² - (1/2)² + (y - (1/2))² - (1/2)²  =  0

(x + (1/2))² + (y - (1/2))²  =  2/4

(x + (1/2))² + (y - (1/2))²  =  1/2

Hence the given is the equation of circle.

(v)  11x² − 25y² − 44x + 50y − 256 = 0

Solution :

11x² − 44x − 25y² + 50y − 256 = 0

A  =  11, B  =  0, C  =  -25

A and C are not equal and they have opposite signs. It is equation of hyperbola.

(vi) y² + 4x + 3y + 4 = 0

Solution :

B  =  0 and A  =  0. Hence it is the equation of parabola.

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