# IDENTIFYING THE CONICS FROM THE GENERAL EQUATION OF THE CONIC

Identifying the Conics from the General Equation of the Conic :

Here we are going to see some practice questions ot identify the conics from the general equation of the conic.

Ax2 + Bxy +Cy2 + Dx + Ey + F = 0.

The graph of the second degree equation is one of a circle, parabola, an ellipse, a hyperbola, a point, an empty set, a single line or a pair of lines. When,

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

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

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

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

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

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

## Identifying the Conics from the General Equation of the Conic - Practice questions

Question 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.

Let us look into the next example on "Identifying the Conics from the General Equation of the Conic".

(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.

Let us look into the next example on "Identifying the Conics from the General Equation of the Conic".

(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.  After having gone through the stuff given above, we hope that the students would have understood, "Identifying the Conics from the General Equation of the Conic".

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