This presentation summarizes the key aspects of the general equation of second degree and how it represents different conic sections. It defines the general equation as ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, and explains how the coefficients relate to the type of conic section represented (ellipse, parabola, hyperbola). It provides examples of reducing specific quadratic equations into standard form and finding the equations of associated geometric elements like the directrix, latus rectum, and axes. The presentation concludes by thanking the audience and inviting questions.

1. Presentation on
Equation of Second Degree
Group: D
Department: Civil Engineering
Dhaka University of Engineering &
Technology, DUET, Gazipur.

2. CONTENTS
 Definition of General Equation of Second Degree
 Conic
 General Equation of Second Degree Represents a Conic
 Applications

3. BACKGROUND
Applications of Ellipse
Applications of Parabola
Applications of Hyperbola

4. General Equation of Second Degree:
Second-degree equation involves at least one variable that is squared or raised to a polar of two. One of the most well-
known second-degree equations is the quadric where a, b, & c are constants and a is not equal to zero. It is represented by
ax2+2hxy+by2+2gx+2fy+c=0
This second-degree equation defines all kinds of figure of Conic.
Conic: If a point P moves in a plane such a way that the ratio of its distance PS from a fixed point S in the plane to its
perpendicular distance PM from a fixed straight line XM in it, is always a constant, the locus of the point P is called a
Conic section or briefly conic.

5. Eccentricity :The constant ratio is called the
eccentricity of the conic and is generally
represented by the letter ‘e’. The fixed point S is
called the focus and the fixed straight line XM is
called directrix of the Conic. ( Fig-1)
The conic is called a Parabola, Ellipse or a
Hyperbola according as the eccentricity e=, < or >,
1
That is Parabola SP=ePM, e=1
For an Ellipse, SP=ePM, e<1
For Hyperbola, SP=ePM, e>1
The eccentricity is a positive number.
M
X
S
P
Fig: 1

6. That is Parabola SP=ePM, e=1
For an Ellipse, SP=ePM, e<1
For Hyperbola, SP=ePM, e>1
The eccentricity is a positive number.

7. The general equation of second degree in x and y represents a conic:
Let the general equation of second degree
ax2+2hxy+by2+2gx+c=0 … … …… …… ..(1)
Let the axes be turned through an angle α, then eq. (1) takes from by putting
xcosα – ysinα for x
and sinα+ycosα for y,
x2(acos2α+2hcosα sinα+bsin2α)+2xy[h(cos2α+sin2α)-(a-b)sinα cosα]+y2(asin2α-2hsinα
cosα+bcos2α)+2x(gcosα+fsinα)+2y(fcosα-gsinα)+c=0 …………..(2)
Now if α be chosen that
h(cos2α-sin2α)-(a-b)sinα cosα=0
or, tan2α=
2ℎ
𝑎−𝑏
then in eq.(2) term containing xy vanishes and it takes the form by putting the value of α
Ax2+By2+2Gx+2Fy+C=0 …………(3)
Where by the principle of invariant
A+B=a+b and AB=ab-h2 ………………(4)

8. Case:1 If A≠0 and B≠0 then the eq.(3) can be written as
A(x+
𝐺
𝐴
)2+B(y+
𝐹
𝐵
)2=
𝐺2
𝐴
+
𝐹2
𝐵
-c ………..(5)
Now shift the origin to the point (
−𝐺
𝐴
,
−𝐹
𝐵
)
Eq.(5) takes the form Ax2+By2=
𝐺2
𝐴
+
𝐹2
𝐵
− c =
Or Ax2+By2=K or (
𝑥2
𝐾
𝐴
+
𝑦2
𝐾
𝐴
)=1…………..(6)
1. If K=0 eq.(6) will represents a pair of straight lines (real of imaginary)
2. If
𝐾
𝐴
and
𝐾
𝐵
in eq.(6) are both positive, then it represents and imaginary ellipse if are both
negative then it represents imaginary ellipse.
3. If
𝐾
𝐴
and
𝐾
𝐵
are of opposite sign eq.(6) will represents a hyperbola

9. Case:II: Let either A or B be zero, then eq.(3) takes the from if A=0
Or, By2+2Gx+2Fy+C=0
Or, B(y+
𝐹
𝐵
)2=-Gx-C+
𝐹2
𝐵
=-2G(x+
𝐵𝐶−𝐹2
𝐵𝐺
)……………..(7)
1. If G=0, then the eq.(7) will represents a pair of parallel straight lines
2. If G=0, and (
𝐹2
𝐵
-C)=0,eq.(7) will represents a pair of coincident lines.
3. G≠0, shift the origin (
𝐵𝐶−𝐹2
𝐵𝐺
,
−𝐹
𝐵
)
then eq.(7) takes the form y2=(
−2𝐺
4
𝑥)
which represents a parabola.

10. Case:(III): When A=B in eq.(3) i.e when a=b and h=0 from 1 ,the eq.(1) will represents a circle.
From the above discussion the general equation of the second degree,
ax2+2hxy+by2+2gx+c=0 will represents
1.A pair of straight lines if the determinant
Δ
𝑎 ℎ 𝑔
ℎ 𝑏 𝑓
𝑔 𝑓 𝑐
=0
Two parallel lines if Δ=0, ab=h2
Two perpendicular lines if Δ=0, a+b=0
2. A circle if a=b, h=0
3. A parabola if ab=h2,Δ≠0
4. An ellipse if ab-h2>0,Δ≠0
5. A hyperbola if ab-h2<0,Δ≠0
Note:Δ=abc+2gh-af2-bg2-ch2=0

11. Question: Reduce the Equation 𝒙 𝟐 + 𝟏𝟐𝒙𝒚 − 𝟒𝒚 𝟐 − 𝟔𝒙 + 𝟒𝒚 + 𝟗 = 𝟎 to the standard
form. Find also the equations of latus rectum, directrices and axes.
Solution:
Let f(x,y)≡ 𝑥2
+ 12𝑥𝑦 − 4𝑦2
− 6𝑥 + 4𝑦 + 9 = 0 ----------(1)
(i)its centre is at (0,
1
2
)
Now transfer the origin to the point(0,
1
2
) ,the equation (1) reduces to
𝑥2
+ 12𝑥𝑦 − 4𝑦2
+ 𝑐 = 0
Where, C=gx1+fy1+c1
= −3.0 + 2.
1
2
+ 9
= 10

12. (ii) The reduced equation is 𝑥2
+ 12𝑥𝑦 − 4𝑦2
+ 10 = 0
or, −
𝑥2
10
−
12𝑥𝑦
10
+
4𝑦2
10
= 1
(iii) The lengths of the semi axes are given by
𝑎 −
1
𝑟2
𝑏 −
1
𝑟2
= ℎ2
Or, −
1
10
−
1
𝑟2
4
10
−
1
𝑟2 = −
6
10
2
Or,𝑟2 =
5
4
, −2
∴ r1
2 =
5
4
,
r2
2 = −2
The conic is a hyperbola.

13. (iv) Equation of the transverse axis is 𝑎 −
1
𝑟1
2 𝑥 + ℎ𝑦 = 0
Or, −
1
10
−
4
5
𝑥 −
3
5𝑦
= 0
Or,3𝑥 + 2𝑦 = 0
Referred to the Centre 0,
1
2
Or, 3(𝑥 − 0) + 2(𝑦 −
1
2
) = 0
Referred to the origin, 3𝑥 + 2𝑦 − 1 = 0
Slope of it, tan 𝜃 = −
3
2
sin 𝜃 =
3
13
cos 𝜃 = −
2
13
Through 0,
1
2
,then k=
3
2
The equation is 4𝑥 − 6𝑦 + 3 = 0

14. (v) Eccentricity,
e2=1 −
𝑟2
2
𝑟1
2
𝑒 =
13
5
(vi) d for foci S and S’ = ±𝑎𝑒
=±𝑟1 ±
13
2
d for feet of the directrices Z and Z’
= ±𝑎𝑒
= ±
5
2 13

15. (vii) Here (h,k) are the Centre 0,
1
2
of conic(1)
Points S and S’= ℎ + 𝑑 cos 𝜃 , 𝑘 + 𝑑 sin 𝜃
= 0 ±
1
2
13 ∗ 2
1
13
,
1
2
±
1
2
13
13. 3
1
13
= −1,2 𝑎𝑛𝑑 1, −1

16. (viii) points Z and Z’
= 0 ±
5
2
1
13
− 2
1
13
,
1
2
±
5
2 13
− 3
1
13
= −
5
13
,
14
13
𝑎𝑛𝑑
5
13
, −
1
13

17. (ix) Latus rectum and directrix are parallel to the conjugate axis 2𝑥 − 3𝑦 +
3
2
= 0
Therefore their slopes are the same as conjugate axis.

18. (x) The equation of the latus rectum is 2x -3y +k = 0.
Since it passes through S(-1,2) or, S’(1,-1)
Hence k = 8 or, - 4
The equations are 3x-2y +8 =0 and 3x-2y -4 =0

19. (xi) Similarly the equation of directrix is 3x-2y+=0
Since it passes through Z −
5
13
,
14
13
𝑜𝑟
5
13
, −
1
13
Hence, =
43
13
, −
17
13
The equation of directrices 3x-2y +4 =0 and 3x-2y -1 =0

20. REFERENCE
 A Text Book On Co-ordinate Geometry With Vector Analysis
By Rahman & Bhattacharjee
 Online Sources

21. Thank You to all

22. Any Questions?