IDENTIFYING THE CONICS FROM THE GENERAL EQUATION OF THE CONIC

General equation of a conic :

Ax2 + Bxy +Cy2 + Dx + Ey + F  =  0

The graph of the second degree equation will be a circle or a parabola or an ellipse or a hyperbola.

(1) When A = C = 1, B = 0, D = -2h, E = −2k, F = h2 + k2 - r2, the general equation reduces to

(x − h)2 + ( y − k)2  =  r2,

which is a circle.

(2) When B = 0 and either A or C = 0 , the general equation yields a parabola under study, at this level.

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

(4) When AC and A and C are of opposite signs, the general equation yields a hyperbola

(5) When A = C and B = D = E = F = 0 , the general equation yields a point x2 + y2 = 0 .

(6) When A = C = F and B = D = E = 0 , the general equation yields an empty set x2 + y2 +1 = 0 , as there is no real solution.

(7) When A  0 or C  0 and others are zeros, the general equation yield coordinate axes.

(8) When A = -C and rests are zero, the general equation yields a pair of lines x2 - y2 = 0

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) =  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

≠ 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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