Exercise 1
Recognise the type of the following conics (1) 2x2+ xy − y2+ x + y − 3 = 0 (2) 2x2+ y2− 3x + y − 5 = 0 (3) x2+ xy − 2y2+ x + 2y = 0
Exercise 1, answer
(1) The complete matrix associated to the conic is
2 12 12
1
2 −1 12
1 2
1
2 −3
whose determinant
6 + 1 8+1
8 +1 4 +3
4−1 2 = 27
4 6= 0 Thus the conic is non-degenerate.
As
2 12
1 2 −1
= −2 −1 4 = −9
4 < 0 the conic is a hyperbola.
Exercise 1, answer (cont.)
(2) The complete matrix associated to the conic is
2 0 −32 0 1 12
−32 12 −5
whose determinant is
−10 −9 4 +3
2 = −43 4 6= 0 Thus the conic is non-degenerate.
As
2 0 0 1
= 2 > 0 the conic is an ellipse.
Exercise 1, answer (cont.)
(3) The complete matrix associated to the conic is
1 12 12
1
2 −2 1
1
2 1 0
whose determinant is
1 4+1
4+1
2 − 1 = 0 Thus the conic is degenerate.
As
1 12
1 2 −2
−9 4 the conic is of hyperbolic type.
Exercise 2
In the plane, let the point A = (3, 1) and the line r : x + y = 0 be given.
Write the equation of the parabola with focus A and directrix r .
Answer. The parabola is the locus of points (x , y ) having same distance from A and from r . This means
(x − 3)2+ (y − 1)2= (x + y )2 1 + 1
2(x2− 6x + 9 + y2− 2y + 1) = x2+ 2xy + y2 x2− 2xy − 12x + y2− 4y + 20 = 0
Exercise 3
In the plane, let the conic C : x2− xy − 2y2− x + 5y = 0 be given.
(1) Recognise the type of the conic and write it in canonical form (2) Write the equation of the parabolae tangent to C in the origin and
passing through the points A = (−1, 0) and B = (0, 1)
Exercise 3, answer
(1) The complete matrix associated with C is
1 −12 −12
−12 −2 52
−12 52 0
whose determinant is
5 8+5
8 +1 2 −25
4 = −9 2 6= 0 Thus the conic is non-degenerate. Since
1 −12
−12 −2
= −2 −1 4 = −9
4 < 0 then C is a hyperbola.
Exercise 3, answer (cont.)
The characteristic polynomial of the incomplete matrix of C is
1 − λ −12
−12 −2 − λ
= λ2+ λ −9 4 so that the eigenvalues are
−1 ±√ 1 + 9
2 = −1 ±√
10 2
The eigenspace associated with the eigenvalue −1+
√ 10
2 is the set of solutions of the equation
3 +√ 10 2 x −1
2y = 0 y = (3 +√
10)x
Thus this space is generated by the vector (1, 3 +√
10), whose norm is p
1 + 9 + 6√
10 + 10 =p
20 + 6√ 10.
Exercise 3, answer (cont.)
The eigenspace associated with the eigenvalue −1+
√ 10
2 is the set of solutions of the equation
3 −√ 10 2 x −1
2y = 0 y = (3 −√
10)x
Thus this space is generated by the vector (1, 3 −√
10), whose norm is p
1 + 9 − 6√
10 + 10 =p
20 − 6√ 10.
Consider the orthogonal matrix
√ 1 20+6√
10
√ 1 20−6√
10 3+√
√ 10 20+6√
10
3−√
√ 10 20−6√
10
Since its determinant is negative, the matrix for performing the change of coordinates is the one obtained by swapping the columns, so that
Exercise 3, answer (cont.)
x y
=
√ 1 20−6√
10
√ 1 20+6√
10 3−√
√ 10 20−6√
10
3+√
√ 10 20+6√
10
X¯ Y¯
x = √ 1
20−6√ 10
X +¯ √ 1
20+6√ 10
Y¯
y = 3−
√
√ 10 20−6√
10
X +¯ 3+
√
√ 10 20+6√
10
Y¯
One obtains the new equation for the conic:
Exercise 3, answer (cont.)
−1 +√ 10 2
X¯2−1 +√ 10 2
Y¯2+
+ 1
p20 − 6√ 10
X −¯ 1
p20 + 6√ 10
Y +¯
+ 15 − 5√ 10 p20 − 6√
10
X +¯ 15 + 5√ 10 p20 + 6√
10 Y = 0,¯
−1 +√ 10 2
X¯2−1 +√ 10 2
Y¯2+ 14 − 5√ 10 p20 − 6√
10
X +¯ 14 + 5√ 10 p20 + 6√
10 Y = 0¯
Exercise 3, answer (cont.)
To get rid of the linear term:
X = X +¯ 14−5
√
√ 10 20−6√
10(√ 10−1)
= ¯X + (14−5
√10)(√ 10+1)
√
20−6√ 10(√
10−1)(√ 10+1)
=
= X +¯ −36+9
√10 9
√
20−6√
10 = ¯X +
√10−4
√
20−6√ 10
Y = Y −¯ 14+5
√
√ 10 20+6√
10(1+√ 10)
= ¯Y − (14+5
√10)(1−√
√ 10) 20+6√
10(1+√ 10)(1−√
10)
=
= Y −¯ −36−9
√10
−9
√
20+6√
10 = ¯Y − 4+
√10
√
20+6√ 10
Exercise 3, answer (cont.)
The canonical form of C is thus
−1 +√ 10
2 X2−1 +√ 10 2 Y2−
−−1 +√ 10 2
26 − 8√ 10 20 − 6√
10+1 +√ 10 2
26 + 8√ 10 20 + 6√
10= 0,
−1 +√ 10
2 X2−1 +√ 10 2 Y2+ +−(−106 + 34√
10)(20 + 6√
10) + (106 + 34√
10)(20 − 6√ 10)
2(400 − 360) = 0
−1 +√ 10
2 X2−1 +√ 10
2 Y2+80 − 44√
10 + 80 + 44√ 10
80 = 0
and, finally,
−1 +√ 10
2 X2−1 +√ 10 2
2
+ 2 = 0
Exercise 3, answer (cont.)
(2) The tangent line to C in the origin is the line −x + 5y = 0. The conics tangent to that line in the origin have equation
ax2+ bxy + cy2− x + 5y = 0
Those passing through A and B are those satisfying the conditions
a + 1 = 0 c + 5 = 0 that is
a = −1, c = −5 Among the conics
−x2+ bxy − 5y2− x + 5y = 0 parabolas are characterised by the condition
5 − b 2
2
= 0 that is
b = ±2√ 5
Exercise 3, answer (cont.)
Consequently, the parabolas have equations
−x2± 2√
5xy − 5y2− x + 5y = 0
Exercise 4
Consider the conic 3x2− 4xy + 8x + 5 = 0. Find the polar of B = (1, 2).
Answer. The polar line of B = (1, 2) is the line ax + by + c = 0 where
a =3 − 4 + 4 = 3 b = − 2
c =4 + 5 = 9 so the equation of the polar is
3x − 2y + 9 = 0
Exercise 5
Determine the tangent line to the conic x2− 2xy + y2− 2x = 0 at the point P = (2, 0).
Answer. The partial derivatives of the left-hand side of the equation are:
∂(x2− 2xy + y2− 2x)
∂x =2x − 2y − 2
∂(x2− 2xy + y2− 2x)
∂y = − 2x + 2y
Computed in P, their values are 2 and −4, respectively. Thus the tangent line has equation
2(x − 2) − 4y = 0 or, equivalently,
x − 2y − 2 = 0
