General Equation of Second Degree
The general equation of second degree 𝑎𝑥 + 2ℎ𝑥𝑦 + 𝑏𝑦 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0 will
represent
(i) a pair of straight line if the determinant
𝑎
Δ= ℎ
𝑔
ℎ
𝑏
𝑓
𝑔
𝑓 =0
𝑐
or, 𝑎𝑏𝑐 + 2𝑓𝑔ℎ − 𝑎𝑓 − 𝑏𝑔 − 𝑐ℎ = 0
(ii) two parallel lines if Δ = 0, 𝑎𝑏 − ℎ = 0
(iii) two perpendicular lines if Δ = 0, 𝑎 + 𝑏 = 0
(iv) a circle if 𝑎 = 𝑏, ℎ = 0
(v) a parabola if 𝑎𝑏 − ℎ = 0, Δ ≠ 0
(vi) a ellipse if 𝑎𝑏 − ℎ > 0, Δ ≠ 0
(vii) a hyperbola if 𝑎𝑏 − ℎ < 0, Δ ≠ 0
(ix) a rectangular hyperbola if 𝑎 + 𝑏 = 0, 𝑎𝑏 − ℎ < 0, Δ ≠ 0
The general equation of second degree
𝑎𝑥 + 2ℎ𝑥𝑦 + 𝑏𝑦 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0
The co-ordinates of centre of (1) is
(1)
,
.
Example: Reduce the equation 16𝑥 − 24𝑥𝑦 + 9𝑦 − 104𝑥 − 172𝑦 + 44 = 0 to the
standard form. Find its all properties.
Solution: Given that 16𝑥 − 24𝑥𝑦 + 9𝑦 − 104𝑥 − 172𝑦 + 44 = 0
(1)
In equation (1) 𝑎 = 16, 𝑏 = 9, 𝑐 = 44, ℎ = −12, 𝑔 = −52, 𝑓 = −86
Δ = 16.9.44 + 2(−86)(−52)(−12) − 16(−86) − 9(−52) − 44(−12) ≠ 0
𝑎𝑏 − (ℎ) = 16.9 − (−12) = 0
Hence equation (1) will represent a parabola.
Equation (1) can be written (4𝑥 − 3𝑦) = 104𝑥 + 172𝑦 − 44
(2)
The lines 4𝑥 − 3𝑦 = 0 and 104𝑥 + 172𝑦 − 44 = 0 are not at right angles, hence let us
introduce a constant 𝜆 in equation (2) then
(4𝑥 − 3𝑦 + 𝜆) = 104𝑥 + 172𝑦 − 44 + 2𝜆(4𝑥 − 3𝑦) + 𝜆
𝑜𝑟, (4𝑥 − 3𝑦 + 𝜆) = 𝑥(104 + 8𝜆) + 𝑦(172 − 6𝜆) + 𝜆 − 44
(3)
The lines 4𝑥 − 3𝑦 + 𝜆 = 0
(4)
and 𝑥(104 + 8𝜆) + 𝑦(172 − 6𝜆) + 𝜆 − 44 = 0
(5)
are at right angles if (formula 𝑎 𝑎 + 𝑏 𝑏 = 0)
4(104 + 8𝜆) + (−3)(172 − 6𝜆) = 0 𝑜𝑟, 𝜆 = 2
(6)
Now equation (3) becomes by (6) (4𝑥 − 3𝑦 + 2) = 40(3𝑥 + 4𝑦 − 1)
which can be written as
4𝑥 − 3𝑦 + 2
√4 + 3
𝑜𝑟,
= 4.2
√
3𝑥 + 4𝑦 − 1
. 25 = 40.5
√4 + 3
or, 𝑌 = 4𝑎𝑋
√
(7)
which is the standard form of a parabola.
Here 𝑌 =
√
, 𝑋=
(8)
√
Now the properties of the above parabola are given below:
(1) Axis of (7) is 𝑌 = 0 𝑜𝑟, 4𝑥 − 3𝑦 + 2 = 0 [by (8)]
(9)
(ii) Tangent at vertex is 𝑋 = 0 𝑜𝑟, 3𝑥 + 4𝑦 − 1 = 0 [by (8)]
(iii) Vertex is intersection of equation (9) and (10)
i.e. solving equation (9) and (10) , then 𝑥 = − , 𝑦 =
so vertex is − ,
(iv) length of latus rectum is 4𝑎 = 4.2 − 8
(v) Fous 𝑋 = 𝑎, 𝑌 = 0 by (8)
𝑋=
3𝑥 + 4𝑦 − 1
√4 + 3
= 2,
4𝑥 − 3𝑦 + 2
√4 + 3
=0
𝑜𝑟, 3𝑥 + 4𝑦 − 11 = 0, 4𝑥 − 3𝑦 + 2 = 0 Solving them
𝑥 = 1, 𝑦 = 2 so, co-ordinate of focus is (1, 2)
(vi) Foot of the directrix, 𝑋 = −𝑎 and 𝑌 = 0
or,
√
= −2,
√
= 0 or, 3𝑥 + 4𝑦 + 9 = 0, 4𝑥 − 3𝑦 + 2 = 0
Solving them, 𝑥 = − , 𝑦 = −
so Foot of the directrix is − . −
(vii) Equation of the directrix is 𝑋 = −𝑎 or, 3𝑥 + 4𝑦 + 9 = 0
(viii) Equation of latus rectum is 𝑋 = 𝑎 or, 3𝑥 + 4𝑦 − 11 = 0
(10)
Example: Reduce the equation 𝑓(𝑥, 𝑦) = 8𝑥 + 4𝑥𝑦 + 5𝑦 − 24𝑥 − 24𝑦 = 0 to the
standard form. Find its all properties.
Solution: Given that 𝑓(𝑥, 𝑦) = 8𝑥 + 4𝑥𝑦 + 5𝑦 − 24𝑥 − 24𝑦 = 0
(1)
In equation (1) 𝑎 = 8, 𝑏 = 5, ℎ = 2, 𝑔 = −12, 𝑓 = −12, 𝑐 = 0
∴ ∆= 𝑎𝑏𝑐 + 2𝑓𝑔ℎ − 𝑎𝑓 − 𝑏𝑔 − 𝑐ℎ = −𝑣𝑒 and 𝑎𝑏 − ℎ == 36
Therefore ∆≠ 0, 𝑎𝑏 − ℎ > 0.
(i) The equation (1) will represent an ellipse.
( , )
= 16𝑥 + 4𝑦 − 24 = 0 and
𝜕𝑓(𝑥, 𝑦)
= 10𝑦 + 4𝑥 − 24 = 0
𝜕𝑦
or, 16𝑥 + 4𝑦 − 24 = 0 and 10𝑦 + 4𝑥 − 24 = 0
Centre of the conic is the intersection of these two lines. Solve, 𝑥 = 1, 𝑦 = 2
(ii) Centre (1,2)
Transfer the origin to a point (1,2) when direction of axes are unchanged, then (1)
becomes
𝑎𝑥 + 2ℎ𝑥𝑦 + 𝑏𝑦 + 𝑐 = 0
i.e. 8𝑥 + 4𝑥𝑦 + 5𝑦 + 𝑐 = 0
where 𝑐 = 𝑥 𝑔 + 𝑦 𝑓 + 𝑐 = −12.1 − 12.2 + 0 = −36
The equation (1) referred to the centre (1,2) is
8𝑥 + 4𝑥𝑦 + 5𝑦 − 36 = 0 or, 8𝑥 + 4𝑥𝑦 + 5𝑦 = 36
or,
𝑥 + 𝑥𝑦 +
where 𝐴 = , 𝐵 =
𝑦 = 1 𝑜𝑟, 𝐴𝑥 + 2𝐻𝑥𝑦 + 𝐵𝑦 = 1
(2)
,𝐻 =
Removing 𝑥𝑦 term when the axes being rotated
∴ 𝑎 𝑥 + 𝑏 𝑦 = 36
(3)
where 𝑎 + 𝑏 = 𝑎 + 𝑏 and 𝑎 𝑏 = 𝑎𝑏 − ℎ
(4) and 𝑎 𝑏 = 8.5 − 4 = 36
𝑎 + 𝑏 = 8 + 5 = 13
Now (𝑎 − 𝑏 ) = (𝑎 + 𝑏 ) − 4𝑎 𝑏 = 169 − 144 = 25
∴ 𝑎 −𝑏 =5
(6) [take +ve sign]
From (4) and (6) we get 𝑎 = 9, 𝑏 = 4
Now equation (3) becomes
9𝑥 + 4𝑦 = 36 or,
+
=1
(5)
or,
+
= 1 which is standard form of an ellipse.
𝐵
𝑍′ 𝐴′
𝑆′
𝐿
A
𝑆
𝐶
𝐵′
𝑍
𝐿′
The length of the semi axes are given by
1
1
− (𝐴 + 𝐵) + 𝐴𝐵 − 𝐻 = 0
𝑟
𝑟
Put the values of 𝐴, 𝐵, 𝐻 from (2) then
𝑟 − 13𝑟 + 36 = 0 𝑜𝑟, 𝑟 = 9 𝑜𝑟 4 𝑖. 𝑒. 𝑟 = 3, 𝑟 = 2
(6)
(iii) The lengths of the axes are 2𝑟 = 6 and 2𝑟 = 4.
The equation of the major and minor axes are
𝐴−
𝑥 + 𝐻𝑦 = 0 and 𝐴 −
𝑥 + 𝐻𝑦 = 0
or, 2𝑥 + 𝑦 = 0 and 𝑥 − 2𝑦 = 0
referred to the centre (1,2),
2(𝑥 − 1) + (𝑦 − 2) = 0 and (𝑥 − 1) − 2(𝑦 − 2) = 0 referred to the old origin.
2𝑥 + 𝑦 − 4 = 0 and 𝑥 − 2𝑦 + 3 = 0
(iv) Equation of the major axis is 2𝑥 + 𝑦 − 4 = 0
(7)
(v) Equation of the minor axis is 𝑥 − 2𝑦 + 3 = 0
(8)
(vi) Eccentricity 𝑒 = 1 −
=1− =
∴ 𝑒=
Now 𝑑 for foci 𝑆 and 𝑆 = ±𝑎𝑒 = ±𝑟 𝑒 = 3.
√
√
(9)
= √5 [by (6) and (9)]
[𝒂 is the semi-major axis]
(vii) Slope of major axis is 𝑡𝑎𝑛𝜃 = −2 and 𝑠𝑖𝑛𝜃 =
√
, 𝑐𝑜𝑠𝜃 = −
√
[from (7)]
[Note: Slope of the axis is given by 𝑡𝑎𝑛𝜃. From it we get 𝑠𝑖𝑛𝜃 and 𝑐𝑜𝑠𝜃. If 𝑡𝑎𝑛𝜃 is
negative, then 𝜃 is obtuse. In this case we keep perpendicular positive and base negative.
Thus 𝑠𝑖𝑛𝜃 is positive and 𝑐𝑜𝑠𝜃 is negative in this case.]
(ℎ, 𝑘) i.e. (1, 2) is the centre of the conic (1).
(viii) Focus 𝑆; (ℎ + 𝑑𝑐𝑜𝑠𝜃, 𝑘 + 𝑑𝑠𝑖𝑛𝜃) = 1 + √5. −
Focus 𝑆′; (ℎ − 𝑑𝑐𝑜𝑠𝜃, 𝑘 − 𝑑𝑠𝑖𝑛𝜃) = 1 − √5. −
√
, 2 + √5.
√
, 2 − √5.
= (0, 4)
√
= (2, 0)
√
𝑑 for vertices 𝐴 and 𝐴′ = ±𝑎 = ±𝑟 = ±3
(ix) Vertex 𝐴; (ℎ + 𝑑𝑐𝑜𝑠𝜃, 𝑘 + 𝑑𝑠𝑖𝑛𝜃) = 1 + 3. −
= 1−
3
√5
,
2+
Vertex 𝐴′; (ℎ + 𝑑𝑐𝑜𝑠𝜃, 𝑘 + 𝑑𝑠𝑖𝑛𝜃) = 1 − 3. −
= 1+
𝑑 for vertices 𝑍 and 𝑍′ = ± = ±
3
√5
=±
.
√
,
√
6
√5
, 2 − 3.
√
2−
=±
, 2 + 3.
√
√
6
√5
√
(x) Foot of directrix 𝑍; (ℎ + 𝑑𝑐𝑜𝑠𝜃, 𝑘 + 𝑑𝑠𝑖𝑛𝜃) = 1 +
. −
√
√
, 2+
√
.
9
18
4 28
= 1− ,2+
= − ,
5
5
5 5
Foot of directrix 𝑍′; (ℎ − 𝑑𝑐𝑜𝑠𝜃, 𝑘 − 𝑑𝑠𝑖𝑛𝜃) = 1 −
√
. −
√
, 2−
√
9
18
14 8
= 1+ ,2−
=
,−
5
5
5
5
𝑑 for the points 𝐵 and 𝐵 = ±𝑏 = ± 𝑟 = ±2 [𝒃 is the semi-minor axis]
Slope of minor axis is 𝑡𝑎𝑛𝜃 = and 𝑠𝑖𝑛𝜃 =
, 𝑐𝑜𝑠𝜃 =
√
√
[from (7)]
(ℎ, 𝑘) i.e. (1, 2) is the centre of the conic (1).
Point 𝐵; (ℎ + 𝑑𝑐𝑜𝑠𝜃, 𝑘 + 𝑑𝑠𝑖𝑛𝜃) = 1 + 2. −
= 1−
4
√5
,2+
√
4
√5
,2−
√
2
√5
Point 𝐵′; (ℎ − 𝑑𝑐𝑜𝑠𝜃, 𝑘 − 𝑑𝑠𝑖𝑛𝜃) = 1 − 2. −
= 1+
, 2 + 2.
√
2
√5
(xi) Length of the latus rectum=
=
=
.
=
, 2 − 2.
√
.
√
√
(xii) Distance between 𝑆 and 𝐿 or 𝐿′ is given by
𝑑=±
𝑏
𝑟
4
=± =±
𝑎
𝑟
3
(ℎ, 𝑘) stands for 𝑆(0,4) 𝑜𝑟 𝑆′(2,0) [by (viii)]
Slope of latus rectum is the same as the slope of minor axis.
Point 𝐿; (ℎ + 𝑑𝑐𝑜𝑠𝜃, 𝑘 + 𝑑𝑠𝑖𝑛𝜃) = 0 + .
Point 𝐿′; (ℎ − 𝑑𝑐𝑜𝑠𝜃, 𝑘 − 𝑑𝑠𝑖𝑛𝜃) = 0 − .
√
√
,4+ .
,4 − .
=
√
√
√
= −
,4 +
√
√
,4−
√
Similarly another pair of points with respect 𝑆′(2,0) can be determined.
(xiii) Equation of the latus rectum which is parallel to minor axis 𝑥 − 2𝑦 + 3 = 0 is
𝑥 − 2𝑦 + 𝑘 = 0. Since it passes through the foci 𝑆(0,4) 𝑜𝑟 𝑆′(2,0), hence 𝑘 = 8, −2.
Hence equations of the latus rectums are 𝑥 − 2𝑦 + 8 = 0 and 𝑥 − 2𝑦 − 2 = 0
(xiv) Equation of the directrix which is parallel to minor axis 𝑥 − 2𝑦 + 3 = 0 is
𝑥 − 2𝑦 + 𝑘′ = 0. Since it passes through 𝑍 − ,
𝑜𝑟
,−
, hence 𝑘′ = 12, −6.
Hence equations of the directrices are 𝑥 − 2𝑦 + 12 = 0 and 𝑥 − 2𝑦 − 6 = 0.
Note: Eccentricity of the conic, 𝑒 is given by 𝑒 = 1 −
If 𝑟 and 𝑟 are positive, the conic is an ellipse. If 𝑟 is negative the conic is a hyperbola. If
𝑟 = 𝑟 and 𝑟 is negative, the conic is a rectangular hyperbola.
Example: Reduce the equation 4𝑥 − 4𝑥𝑦 + 𝑦 − 8𝑥 − 𝑦 + 6 = 0 to the standard form.
Find its all properties.