..READ THIS FIRST..About a circle..About a
Parabola..About an
ellipse..About a
Hyperbola..About
homogeneous coordinates and ideal points..About
imaginary points and lines..About
degenerated conic sections..About a
tangent lines..About
asymptotes..About
systems of conic sections..About pole
and polar line..About
center-point of a conic section..About
center-line of a conic section..About axes
of an affine conic section..About foci
of a conic section..About the
locus of a point
READ THIS
FIRST
If a problem is solved. It is not 'the' answer.No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.
About a circle
Level 2 problemsA variable circle c has equation
x2 + y2 - 2 (t2 - 3 t + 1) x - 2 (t2 + 2 t) y + t = 0
The number t is a parameter. Calculate the point P with a constant
power with respect to c. How much is that power.
|
Say P has coordinates (r,s), then the power of P with respect to c is
r2 + s2 - 2 (t2 - 3 t + 1) r - 2 (t2 + 2 t) s + t
<=> -(2 s + 2 r) t2 + (6 r - 4 s + 1) t + r2 + s2 - 2 r
This power is independent of the parameter t if and only if
2 s + 2 r = 0 and 6 r - 4 s + 1= 0
<=> r = 1/10 and s = -1/10
The point P(0.1; -0.1) has a constant power with respect to the variable
circle. This power is 0.22 .
About a Parabola
Level 2 problems-
The parabola P has equation y2 = 2 p x.
A variable point P has coordinates (p/2,t). The parameter is the real number t greater than p .
Calculate the the tangent of the sharp angle between the tangent lines through P.
From the theory about the tangent lines through a given point P(xo,yo), we know that the slopes are given by
______________ | 2 yo + \| yo - 2 p xo m1 = --------------------- 2 xo ______________ | 2 yo - \| yo - 2 p xo m2 = --------------------- 2 xoIn our case here, xo = p/2 and yo = t. So, the slopes are______________ | 2 2 t + \| t - p m1 = --------------------- p ______________ | 2 2 t - \| t - p m2 = --------------------- p The tangent of the angle between the lines is given by m1 - m2 ----------- 1 + m1 m2 Now, __________ | 2 2 2 \| t - p m1 - m2 = -------------- and m1.m2 = 1 p Then the tangent of the angle between the lines is __________ | 2 2 \| t - p ------------- p -
The tangent lines t and t' in points P(x1,y1) and in P'(x2,y2) of a parabola are orthogonal.
Prove that y1.y2 = - p.p
From the theory about the tangent line in a given point P(xo,yo) of a parabola, we know that the slope is p/yo .
Thus, in point P is the slope p/y1 and in point P' is the slope p/y2.
The tangent lines are orthogonal if and only ifp p -- . -- = -1 y1 y2 <=> y1.y2 = - p.p -
Point D(xo,yo) is on the parabola P has equation y² = 2 p x.
The normal through point D meets the x-axis in point N.
The orthogonal projection of D on the x-axis is E.
Prove that the distance |E,N| is constant. The vector EN is calles the constant subnormal of the parabola.
From the theory about the tangent line in a given point D(xo,yo) of a parabola, we know that the slope is p/yo .
The slope of the normal is -yo/p.
The normal line has equation (y-yo) = (-yo/p)(x-xo).
The coordinates of N are (xo + p,0).
The coordinates of E are (xo,0).
The distance |E,N| is p.
About an ellipse
Level 1 problems-
Calculate the equation of the normal in a point P(xo,yo) of the ellipse b2 x2 + a2 y2 = a2 b2
We know that the tangent line in a point P(xo,yo) isa2 yo y + b2 xo x = a2 b2 the slope of this line is b2 xo - ----- a2 yo the slope of the normal is a2 yo ----- b2 xo The normal is a2 yo y - yo = ------- (x - xo) b2 xo
-
Calculate the product of the distances from the foci of an ellipse to the tangent line.
Take the ellipse E
b2 x2 + a2 y2 = a2 b2 and a point D(a cos(t) , b sin(t)) on that ellipse The tangent line in D is a2 yo y + b2 xo x = a2 b2 <=> a2 b sin(t) y + b2 a cos(t) x = a2 b2 <=> a sin(t) y + b cos(t) x = a b The normal equation of that tangent line is a sin(t) y + b cos(t) x - a b ------------------------------ = 0 _________________________ | 2 2 2 2 \| a sin (t) + b cos (t) The distance from f(c,0) to that line is b cos(t) c - a b | ------------------------------| _________________________ | 2 2 2 2 \| a sin (t) + b cos (t) The distance from f'(-c,0) to that line is - b cos(t) c - a b | ------------------------------| _________________________ | 2 2 2 2 \| a sin (t) + b cos (t) The product of the distances is a2 b2 - b2 c2 cos2(t) | ---------------------------| a2 sin2(t) + b2 cos2(t) b2(a2- c2 cos2(t)) = | ------------------------------ | a2 (1 - cos2(t)) + b2 cos2(t) b2(a2- c2 cos2(t)) = | --------------------| a2 - c2 cos2(t) = b2
-
Take all chords, with slope m, of an ellipse. Show that all midpoints of these chords are on one line.
Take the ellipse E2 2 x y -- + -- = 1 2 2 a b <=> b2 x2 + a2 y2 = a2 b2and the variable line y = m x + t with slope m. (t = parameter)
The intersection point of the ellipse and the line are the solutions of the system/ | b2 x2 + a2 y2 = a2 b2 | \ y = m x + tSubstitution givesb2 x2 + a2 (m x + t)2 - a2 b2 = 0 <=> (b2 + m2 a2) x2 + 2 a2 t m x + t2 a2 - b2 a2 = 0Say x1 and x2 are the roots of this equation. These are the first coordinates of the intersection points.
The first coordinate of the midpoint if these points isx1 + x2 - a2 t m ------- = ----------- 2 b2 + m2 a2 The variable midpoint of the chord is the intersection of the lines - a2 t m x = ------------ and y = m x + t b2 + m2 a2The elimination theory says that we can find the equation of the locus of the midpoint of the chord by eliminating the parameter t between previous equations.
Substituting t from the second in the first equation givesm y a2 + x b2= 0 <=> b2 x y = - ---- a2 mAll the midpoints of the chords are on this line.
About a Hyperbola
Level 1 problems-
A hyperbola H has vertices P and P' an D is on H.
Prove that the product of the slopes of DP and DP' is constant.
Take D(a sec(t) , b tan(t)) on H.
The slope of DP isb tan(t) ------------- a(sec(t)- 1)The slope of DP'isb tan(t) ------------- a(sec(t)+ 1)The product of the slopes of DP and DP' isb2tan2(t) b2 --------------- = ----- a2(sec2(t)- 1) a2 -
Calculate the lines with slope = 1 and tangent to the hyperbola 9 x2 - 25 y2 = 225
Find the equation of the line connecting the points of tangency.
Asw: y = x + 4 ; y = x - 4 ; 9x - 25 y = 0
-
A point P(xo,yo) is on the hyperbola H with foci F and F'.
Prove that |D,F| = | a - (c/a) xo |.
Take xo = a sec(t) and yo = b tan(t) . Then P(xo,yo) is on the hyperbola H .
|D,F|2= (c - a sec(t))2 + b2 tan2(t) = (c - a sec(t))2 + (c2 - a2) (sec2(t) - 1) = ... = a2 - 2 a c sec(t) + c2 sec2(t) = (a - c sec(t))2 c xo = (a - ----)2 a -
Calculate the lines with slope = m and tangent to the hyperbola b2 x2 - a2 y2 = a2 b2
____________ ____________ | 2 2 2 | 2 2 2 y = m x + \| a m - b and y = m x - \| a m - b -
Prove that the slopes m of the lines through P(xo,yo) and tangent to the hyperbola b2 x2 - a2 y2 = a2 b2 are the solutions of the equation (xo2 - a2) m2 - 2 xo yo m + yo2 + b2 = 0
-
The tangent line in point P of a hyperbola has with the asymptotes the intersection points Q and Q'.
Show that P is the midpoint of the segment [Q,Q'].
About homogeneous coordinates and ideal points
Level 1 problems-
Give a triple of homogeneous coordinates of the points with cartesian coordinates (-3,5);(0,1);(3,0);(0,0). (-3,5) becomes (-3,5,1) or (-30,50,10) or ... (0,1) becomes (0,1,1) or (0,10,10) or ... (3,0) becomes (3,0,1) or (30,0,10) or ... (0,0) becomes (0,0,1) or (0,0,10) or ...
-
Give the cartesion coordinates of the point with homogeneous coordinates (5,2,4);(0,0,7);(1,2,0). (5,2,4) becomes (5/4,1/2) (0,0,7) becomes (0,0) (1,2,0) impossible
-
Give line coordinates, homogeneous equation and the ideal point of the lines 3 x + 5 y - 7 = 0; 24 x = 0; x - y = 0
3 x + 5 y - 7 = 0 gives (3,5,-7) 3 x + 5 y - 7 z = 0 and (-5,3,0) 24 x = 0 gives (1,0,0) 24 x = 0 and (0,1,0) x - y = 0 gives (1,-1,0) x - y = 0 and (1,1,0)
-
Give the homogeneous equation of the line PQ with P(1,4) and Q the ideal point of the line 2 x + y - 4 = 0.
Point Q is (1,-2,0). So, the equation of PQ is
| x y z | | 1 -2 0 | = 0 <=> -2 x - y + 6 z = 0 | 1 4 1 | -
Determine m such that the points (4,1,2);(1,m,5);(2,5,6) are collinear.
The condition is
|4 1 2 | 43 |1 m 5 | = 0 <=> m = -- |2 5 6 | 10 -
Give the coordinates of a variable point of the line 2 x + y - 4 = 0.
We choose two simple points on the line. P(0,4,1) and Q(1,-2,0). A variable point has coordinates
with homogeneous parameters (k x1 + l x2, k y1 + l y2, k z1 + l z2) (l,4k-2l,k) with non-homogeneous parameters (x1 + h x2, y1 + h y2, z1 + h z2) (h,4 - 2h,1) -
Give the line l through the intersection point of 2 x + y - 4 z = 0 and 3 x + 5 y - 7 z = 0 and such that the ideal point (1,2,0) is on line l.
A variable line through the intersection point of 2 x + y - 4 z = 0 and 3 x + 5 y - 7 z = 0 has equation
(2 x + y - 4 z) + h(3 x + 5 y - 7 z) = 0 The given ideal point is on that line <=> 4 + 13 h = 0 <=> h = -4/13 The line l is 14 7 24 -- x - -- y - -- z = 0 <=> 14 x - 7 y - 24 z = 0 13 13 13 -
Calculate the intersection point of the lines 2 x + y - 4 z = 0 and 3 x + 5 y - 7 z = 0. | 1 -4 | | 2 -4 | | 2 1| ( | 5 -7 | , - | 3 -7 | ,| 3 5| ) <=> (13 , 2, 7) -
Calculate the midpoint of (5,7,8) and (4,-6,1)
The cartesian coordinates are
(5/8,7/8) and (4,-6) The midpoint is 37 41 (--, - --) or (37,-41,16) 16 16
About imaginary points and lines
Level 1 problems-
Are the following lines k and l conjugate imaginary lines? line k has line coordinates (1-i,i,4) line l has line coordinates (2,-1-i,4-4i)The lines k and l are conjugate imaginary lines <=> The lines (1+i,-i,4) and (2,-1-i,4-4i) are coinciding <=> (1+i,-i,4) and (2,-1-i,4-4i) are proportionalSince (2,-1-i,4-4i) = (1-i).(1+i,-i,4), the lines k and l are conjugate imaginary lines. -
Calculate three imaginary points on the line x - 2 y + z = 0
Choose two real points (2,1,0) and (1,0,-1).
(2,1,0) + r.i.(1,0,-1) are imaginary points on the line.
For r = 1, 2, 3 we have the points
(2 + i, 1, -i), (2 + 2 i, 1, -2 i), (2 + 3 i, 1, -3 i) -
Calculate the real point on the line k(2 + i, 1, -i)
The real point is the intersection point ofthe lines k(2 + i, 1, -i) and l(2 - i, 1, i). This point is ( i, -4 i, 2 i) or (1, -2, 1)
Level 2 problems
-
Calculate the real values m and n such that the point (2 - i, 3 - n i, m + i)is a real point.
(2 - i, 3 - n i, m + i) is a real point <=> There are real values b and c such that 3 - n i = b (2 - i) and m + i = c (2 - i) <=> b = 3/2; n = 3/2; m = -2; c = -1 Then the point is (2 - i, 3 - 3/2 i, -2 + i) <=> (2 - i, (3/2)(2 - i), -1(2 - i)) <=> (1, 3/2, -1) <=> (2, 3, -2)
About degenerated conic sections
Level 1 problems-
Calculate the components of the curve x2 + 4 x - 6 = 0x2 + 4 x - 6 = 0 <=> ____ ____ x = V 10 - 2 or x = - V 10 - 2These are two lines parallel to the y-axis. -
Calculate the components of the curve x2 + 4 x y + 3 y2 - 2 x z - 4 y z + z2= 0
To factorize this expression, we consider it as a quadratic equation of x. Collecting terms involving x, we have:
x2 + (4 y - 2 z) x + (3 y2 - 4 z y + z2) = 0 The discriminant is (4 y - 2 z)2 - 4 (3 y2 - 4 z y + z2) = 4 y2 The roots are x = z - y, and x = z - 3 y The two components are x + y - z = 0 and x + 3 y - z = 0 -
Calculate the real values of m such that the following conic section is degenerated. x2 - 4 x y + 2 y2 - 5 y - m x + 2 = 0
The homogeneous equation is
x2 - 4 x y + 2 y2 - 5 y z - m x z + 2 z2= 0 The conic section is degenerated. <=> DELTA = 0 <=> | 1 -2 -m/2 | | -2 2 -5/2 | = 0 | -m/2 -5/2 2 | <=> 2 m2 + 20 m + 41 = 0 <=> 3 ___ 3 ___ m = - V 2 - 5 or m = - - V 2 - 5 2 2 -
Calculate the double points of the following conic section x2 + 4 x y + 3 y2 - 2 x z - 4 y z + z2= 0The double point is the solution of the system 2 x + 4 y - 2 z = 0 4 x + 6 y - 4 z = 0 -2 x - 4 y + 2 z = 0 The solution is x = 1, y = 0, z = 1 -
Calculate the double points of the following conic section x2 + 2 x y + y2 - 8 x z - 8 y z + 16 z2= 0The double point is the solution of the system x + y - 4 z = 0 x + y - 4 z = 0 -4 x - 4 y + 16 z = 0 The system is equivalent with the single equation x + y - 4 z = 0All the solutions of this equation are double points.
The conic section is a degenerated parabola. It consists of two coinciding lines. The equation can be written as(x + y - 4 z)2 = 0
About a tangent lines
Level 1 problems-
Calculate the tangent line in point P(2,0) of the conic section x2 - 4 x y - y2 + 2 x - 4 y - 8 = 0The equation is 2(2 x - 4 y + 2 z)+0(-4 x - 2 y - 4 z)+1(2 x - 4 y - 16 z) = 0 <=> x - 2 y - 2 z = 0 -
Calculate the tangent line in point P(2,0) of the conic section x2 + x y - 6 y2 - 4 x z - 2 y z + 4 z2= 0The formule gives 2(2 x + y - 4 z) + 0(x - 12 y - 2 z) + 1(-4 x - 2 y + 8 z) =0 <=> 0 = 0 this is not the equation of a line.Since Fx'(2,0,1) = 0 and Fy'(2,0,1) = 0 and Fz'(2,0,1) = 0 the point P is a double point and each line through P is a tangent line.
-
The tangent line in point P(?,?) of the conic section x2 - 4 x y - y2 + 2 x - 4 y - 8 = 0 is x - 2 y - 2 = 0 Calculate the coordinates of the tangent point.
- method 1.
The conic section and the line intersect in the tangent point. The coordinates are the solution of the system/ x2 - 4 x y - y2 + 2 x - 4 y - 8 = 0 \ x - 2 y - 2 = 0Substituting x from the second equation in the first one, we find y = 0 and from this x = 2. Point P is (2,0). - method 2.
The tangent point is the intersection point of the tangent line and the tangent chord of a simple point on the tangent line. Take Q(0,-1) on the tangent line.
The tangent chord is -x + 2 y + 2 = 0.
The tangent point is the solution of the system/ -x + 2 y + 2 = 0 \ x - 2 y - 2 = 0The solution is P(2,0). - method 3.
Say P(xo,yo,1) is the tangent point. The tangent line in P has equationx(2 xo - 4 yo + 2)+y(-4 xo - 2 yo - 4)+z(2 xo - 4 yo - 16)=0That line is the same line as x - 2 y - 2 = 0. The corresponding coefficients must be directly propertional.
So, xo and yo are the solutions of the system/ 2 xo - 4 yo + 2 = r | -4 xo - 2 yo - 4 = -2r \ 2 xo - 4 yo - 16 = -2rThe solution is xo = 2, yo = 0, r = 6.
- method 1.
About asymptotes
Level 1 problems-
Calculate the asymptotes of the conic section x2 - 4 x y - y2 + 2 x - 4 y - 8 = 0
The slopes m corresponding with the ideal points are the solutions of- m2 - 8 m - 5 = 0 <=> ____ ____ m1 = V 11 - 4 and m2 = -4 - V 11 The asymptotes are the tangent lines in the ideal points ____ ____ P1(1, V 11 - 4,0) and P2(1, -4 - V 11 ,0) The asymptotes are ____ (2 x - 4 y + 2 z) + (V 11 - 4)(-4 x - 2 y - 4 z) = 0 and ____ (2 x - 4 y + 2 z) + (-V 11 - 4)(-4 x - 2 y - 4 z) = 0 -
Calculate the asymptotes of the conic section x2 + 2 x y + y2 - 4 x - 5 y + 7 = 0
Since delta = 0, the conic section has two coinciding ideal points.
Since DELTA is not zero, that ideal point can't be a double point.
Therefore, the tangent line in that ideal point is the ideal line. The asymptotes are z = 0 and z = 0. -
Calculate the asymptotes of the conic section x2 + 2 x y - 4 x = 0
This conic section is degenerated in the linesx = 0 and x + 2 y - 4 = 0 .These lines are the asymptotes because these lines are the tangent lines in the ideal points of te conic section. -
Calculate the asymptotes of the conic section x2- 2 x = 0
The conic section is a degenerated parabola.
The ideal point (0,1,0) is a double point of the degenerated conic section. Each line through this point is an asymptote. So this conic section has an infinity number of asymptotes. All these asymptotes are parallel. -
Search the equation of the conic section with asymptotes x = 0 and x + 2 y - 41 = 0 and through the point P(2,1).The conic section has an equation of the form x (x + 2 y - 41) + k = 0 The point P(2,1) is on the conic section if and only if 2 (2 + 2 - 41) + k = 0 <=> k = 74 The conic section has equation x (x + 2 y - 41) + 74 = 0 <=> x2 + 2 x y - 41 x + 74 = 0
About systems of conic sections
Level 1 problems-
Calculate the equation of the system of conic sections with basic points A(1,2); B(2,0); C(-1,1); D(0,3). Line AB: 2 x + y - 4 z = 0 Line CD: -2 x + y - 3 z = 0
The lines AB and CD form a degenerated conic section of the system. We choose it as the first basic conic section. The equation of this conic section is(2 x + y - 4 z)(-2 x + y - 3 z) = 0 <=> -4 x2 + y2 + 2 x z - 7 y z + 12 z2= 0 Line AC: x - 2 y + 3 z = 0 Line BD: -3 x - 2 y + 6 z = 0The lines AC and BD form a degenerated conic section of the system. We choose it as the second basic conic section. The equation of this conic section is-3 x2 + 4 x y + 4 y2 - 3 x z - 18 y z + 18 z2= 0The equation of the system of conic sections with basic points A(1,2); B(2,0); C(-1,1); D(0,3) isk(-4 x2 + y2 + 2 x z - 7 y z + 12 z2) + l( -3 x2 + 4 x y + 4 y2 - 3 x z - 18 y z + 18 z2) = 0k and l are homogeneous parameters.
With non-homogeneous parameter h, we have the equation:(-4 x2 + y2 + 2 x z - 7 y z + 12 z2) + h( -3 x2 + 4 x y + 4 y2 - 3 x z - 18 y z + 18 z2) = 0 -
Calculate the equation of the system of conic sections with basic points A(1,2); B(2,0); C(-1,1); C(-1,1) and such that the line c with equation x + y = 0 is the tangent line in point C.
Degenerated conic sections of the system are line AB with line c, and AC with line BC. The equations of these degenerated conic sections are
(2 x + y - 4 z)(x + y) = 0 <=> 2 x2 + 3 x y + y2 - 4 x z - 4 y z = 0 and (x - 2 y + 3 z)(-x - 3 y + 2 z) = 0 <=> - x2 - x y + 6 y2 - x z - 13 y z + 6 z2= 0The system has equation:k(2 x2 + 3 x y + y2 - 4 x z - 4 y z) + l(- x2 - x y + 6 y2 - x z - 13 y z + 6 z2) = 0 -
Calculate the equation of the system of conic sections with basic points A(1,2); A(1,2); B(2,0); B(2,0) and such that the line a with equation x + y - 3 z = 0 is the tangent line in point A and b with equation x + 2 y - 2 z = 0 is the tangent line in point B.
Degenerated conic sections of the system are line a with line b, and line AB with line AB. The equations of these degenerated conic sections are
(x + y - 3 z)(x + 2 y - 2 z) = 0 and (2 x + y - 4 z)2= 0The system has equation:k(x + y - 3 z)(x + 2 y - 2 z) + l(2 x + y - 4 z)2= 0 -
Calculate the basic points of the system with basic conic sections x2 + 2 x y + 7 y2 - 5 x z - 17 y z + 6 z2= 0 and -3 x2 - 4 x y + 5 y2 + 3 x z - 9 y z + 6 z2= 0
The basic points are the solutions of the system
/ | x2 + 2 x y + 7 y2 - 5 x z - 17 y z + 6 z2= 0 | | -3 x2 - 4 x y + 5 y2 + 3 x z - 9 y z + 6 z2= 0 \To calculate these points we first calculate a value of r such that(x2 + 2 x y + 7 y2 - 5 x z - 17 y z + 6 z2) + r(-3 x2 - 4 x y + 5 y2 + 3 x z - 9 y z + 6 z2) = 0is a degenerated conic section.| 2 - 6 r 2 - 4 r 3 r - 5 | DELTA = | 2 - 4 r 14 + 10 r -17 - 9 r | | 3 r - 5 -17 - 9 r 12 + 12 r | = -300 (r + 1) (r - 1)2The conic section is degenerated for r = -1 and r = 1.
We take r = 1. Then the degenerated conic section is-2 x2 - 2 x y + 12 y2 - 2 x z - 26 y z + 12 z2= 0 <=> (x - 2y + 3 z)(x +3 y - 2 z) = 0The basic points are the solutions of the systems/ | (x - 2y + 3 z) = 0 | | -3 x2 - 4 x y + 5 y2 + 3 x z - 9 y z + 6 z2 \ and / | (x +3 y - 2 z) = 0 | | -3 x2 - 4 x y + 5 y2 + 3 x z - 9 y z + 6 z2 \The first system has solutions (-1,1,1) and (1,2,1).
The second system has solutions (2,0,1) and (-1,1,1).
These points are the four basic points of the system.
About pole and polar line
Level 1 problems-
Calculate the polar line of P(1,1,1) with respect to the conic section -3 x2 - 4 x y + 5 y2 + 3 x z - 9 y z + 6 z2= 0The polar line is 1.Fx' (x,y,z) + 1.Fy' (x,y,z) + 1.Fz' (x,y,z) = 0 <=> -7 x - 3 y + 6 z = 0 -
Calculate the polar line of P(1,1,1) with respect to the conic section x2 - y2 - 2 x z + 2 y z = 01.Fx' (x,y,z) + 1.Fy' (x,y,z) + 1.Fz' (x,y,z) = 0 <=> 2 x - 2 z + 2 z - 2 y + 2 y - 2 x = 0 <=> 0 = 0This method gives no result because the point P(1,1,1) is a double point of the conic section. Each line is a polar line of a double point.
-
Calculate the polar line p of P(1,0,1) with respect to the conic section (x - y) (x + y - 2 z) = 0This conic section has double point S(1,1,1).
Show that the components of the conic section, the line SP and the line p form a harmonic quartet of lines.
The polar line is
1.Fx' (x,y,z) + 0.Fy' (x,y,z) + 1.Fz' (x,y,z) = 0 <=> 2 y - 2 z = 0 <=> y - z = 0The line SP is x - z = 0.The equations of the four lines can be written as
x - z = 0 y - z = 0 (x - z) - 1.(y - z) = 0 (parameter h = 1) (x - z) + 1.(y - z) = 0 (parameter h' = - 1)Since h = -h' , we have a harmonic quartet of lines. -
Calculate the point C of a polar triangle ABC of the conic section x2 + 2 x y + 7 y2 - 5 x z - 17 y z + 6 z2= 0 if you know that A(2,1,1) and B(0,15,1).
The point C is the intersection point of the polar lines of A and B. So, it is the solution of the system/ x + y - 15 z = 0 | \ 25 x + 193 y - 243 z = 0The coordinates of C are (221,-11,14). -
The lines a,b and c are the polar lines of the vertices of a triangle ABC, with respect to a not degenerated conic section K. P is the intersection point of a and BC.
Q is the intersection point of b and CA.
R is the intersection point of c and AB.Prove that P,Q and R are collinear.
Since all properties in this problem are projective, we can solve it in the projective plane.Choose A(1,0,0) B(0,1,0) and C(0,0,1) Let K : a x2 + 2 b" x y + a' y2 + 2 b' x z + 2 b y z + a" z2= 0 Then line a has equation a x + b" y + b' z = 0 line b has equation b" x + a' y + b z = 0 line c has equation b' x + b y + a" z = 0 and P(0,b',-b") Q(b,0,-b") R(b,-b',0) Since | 0 b' -b"| | b 0 -b"| = 0 | b -b' 0 | the points P, Q and R are collinear.
About center-point of a conic section
Level 1 problems-
Calculate the center-point of the conic section x2 + 2 x y + 7 y2 - 5 x z - 17 y z + 6 z2= 0
The coordinates of the center-points are the solutions of the system/ Fx' (x,y,z) = 0 \ Fy' (x,y,z) = 0 <=> / 2 x + 2 y - 5 z = 0 \ 2 x + 14 y - 17 z = 0 <=> x = 3 ; y = 2 ; z = 2The center point is (3,2,2). -
Calculate the center-point of the conic section x2 + 4 x y + 4 y2 + 2 x z + 4 y z - 8 z2 = 0
The coordinates of the center-points are the solutions of the system/ Fx' (x,y,z) = 0 \ Fy' (x,y,z) = 0 <=> / 2 x + 4 y + 2 z = 0 \ 4 x + 8 y + 4 z = 0All the points of the line x + 2 y + z = 0 are center-points of the degenerated parabola. -
What is the general equation of a conic section with the point (0,0,1) as a center-point.
A conic section has an equation of the forma x2 + 2 b" x y + a' y2 + 2 b' x z + 2 b y z + a" z2= 0
The coordinates of the center-points are the solutions of the system/ Fx' (x,y,z) = 0 \ Fy' (x,y,z) = 0 <=> / a x + b" y + b' z = 0 \ b" x + a' y + b z = 0 (0,0,1) is a solution of this system <=> b' = b = 0The general equation of a conic section with the point (0,0,1) as a center-point isa x2 + 2 b" x y + a' y2 + a" z2= 0
About center-line of a conic section
Level 1 problems-
Calculate the center-line of the conic section x2 + 2 x y + 7 y2 - 5 x z - 17 y z + 6 z2= 0 conjugated to the direction with slope -1.
This center line is the polar line of the point (1,-1,0).
This line has equationFx' (x,y,z) - Fy' (x,y,z) = 0 <=> 2 x + 2 y - 5 z - (2 x + 14 y - 17 z) = 0 <=> z - y = 0So, the center-line is the line y = 1. -
Calculate the value of r such that the line x + 2 y + z = 0 is a center-line of the conic section x2 + 2 x y - y2 - r x z + r y z = 0The center-line of the point (1,m,0) is Fx' (x,y,z) + m Fy' (x,y,z) = 0 <=> 2 x + 2 y - r z + (2 x - 2 y + r z) m = 0 <=> (2 m + 2) x + (2 - 2 m) y + (r m - r) z = 0 This line is the line x + 2 y + z = 0 <=> (2 m + 2) (2 - 2 m) (r m - r) --------- = ----------- = ----------- 1 2 1 <=> 1 m = - -, r = -1 3 -
Calculate the value of r and s such that the line x + 2 y + z = 0 is a center-line conjugated to the direction with slope -2 with respect to the conic section x2 + 2 x y - y2 - r x z + s y z -2 z2 = 0The center-line of the point (1,-2,0) is Fx' (x,y,z) - 2 Fy' (x,y,z) = 0 <=> 2 x + 2 y - r z - 2 (2 x - 2 y + s z) = 0 <=> -2 x + 6 y - (2 s + r) z = 0 We see that there are no values of r and s such that the line x + 2 y + z = 0 coincide with the line -2 x + 6 y - (2 s + r) z = 0 -
Calculate the direction conjugated to (1,-2,0) with respect to the conic section x2 + 2 x y - y2 - 4 x z + 2 y z -2 z2 = 0(1,-2,0) and (1,m,0) are conjugated directions <=> a r1 r2 + b"(r1 s2 + s1 r2) + a' s1 s2 = 0 <=> 1.1.1 + 1.(1 .m - 2 . 1) - 1 .(-2).m = 0 <=> m = 1/3The directions (1,-2,0) and (1,1/3,0) are conjugated. -
Calculate the center-line conjugated to the center-line 4 x - 2 y + z = 0 with respect to the conic section x2 + 2 x y - y2 - 4 x z + 2 y z -2 z2 = 0
The center-line conjugated to the center-line 4 x - 2 y + z = 0, is the center line conjugated to the direction of 4 x - 2 y + z = 0. So, we calculate the center-line conjugated to (1,2,0).
This line is 3 x - y = 0.
About axes of an affine conic section
Level 1 problems-
Calculate the axes and the vertices of the conic section x2 + 2 x y - (1/2) y2 - 4 x z + 2 y z -6 z2 = 0
First we calculate the main directions(1,m,0) is a main direction <=> b" + (a' - a) m - b" m2 = 0 <=> 1 + (-1/2 - 1) m - 1 m2 = 0 <=> m = -2 ; m = 1/2The axes are the center-lines conjugated to these directions.2 y - x - 4 z = 0 and 2 x + y - 2 z = 0The vertices on the axis 2 y - x - 4 z = 0 are the solutions of the system/ 2 y - x - 4 z = 0 | \ x2 + 2 x y - (1/2) y2 - 4 x z + 2 y z -6 z2 = 0 We choose z = 1 / 2 y - x - 4 = 0 | | x2 + 2 x y - (1/2) y2 - 4 x + 2 y - 6 = 0 \ The vertices are the points 4 ___ 2 ____ (- -- V 30 , 2 - -- V 30 ) 15 15 4 ___ 2 ____ (-- V 30 , 2 + -- V 30 ) 15 15Similarly, you'll find the vertices on the other axis -
Calculate the axis of the parabola x2 + 2 x y + y2 - 4 x + 2 y - 6 = 0
The ideal point of the parabola is (1,-1,0). The orthogonal direction is (1,1,0). The axis of the parabola is the polar line of (1,1,0).
It is the line 2 x + 2 y - 1 = 0.
About foci of a conic section
Level 1 problems-
Calculate the focus and the directrix of the parabola x2 + 2 x y + y2 - 4 x + 2 y - 6 = 0
The focus of the parabola is the intersection point of the two Plucker lines. These Plucker lines have equation/ | (Fx' (x,y,1))2 - (Fy' (x,y,1))2 = 4(a - a') F(x,y,1) | | | Fx' (x,y,1).Fy' (x,y,1) = 4 b" F(x,y,1) \ <=> / 2 y + 2 x - 1 = 0 | \ 3 x - 3 y + 4 = 0 <=> 5 11 x = - --, y = -- 12 12 The focus is (-5,11,12)The directrix is the polar line of the focus. This directrix is-5 Fx' (x,y,1) + 11 Fy' (x,y,1) + 12 Fz' (x,y,1) = 0 <=> -6 x + 6 y + -17 z = 0 -
The origin point (0,0,1) is the focus of a not degenerated parabola. The distance from the origin to the vertex of the parabola is r. Show that the distance from the origin to the directrix is 2r.
First method:
Choose the x-axis on the axis of the parabola and the coordinates of the vertex (-r,0,1). Now the equation of the parabola isy2 = 4 r (x + r) <=> y2 = 4 r (x z + r z2) <=> y2 - 4 r x z - 4 r2 z2= 0 The directrix is the polar line of the origin and has equation -4 r x - 8 r2 z = 0 <=> x + 2 r z = 0 <=> x = - 2 rIt is clear that the distance from the origin to this directrix is 2r.Second method:
Since the vertex of each parabola is the midpoint between the focus and the directrix and since the origin is the focus, the distance from the origin to the directrix is 2r.
About the locus of a point
Level 1 problems-
Given is a quadratic equation in z with parameters x and y. z2 - x z + (x - y)2 = 0The parameters x and y are the coordinates of a point P with respect to an orthonormal coordinate system in a plane.Calculate the locus of point P such that the quadratic equation has equal roots.
The quadratic equation has equal roots <=> The discriminant D = 0 <=> x2 - 4(x - y)2 = 0 <=> -3 x2 + 8 x y - 4 y2 = 0This is the equation of the locus of P. It is the quadratic equation of two lines through the origin of the coordinate system. The slopes of the lines are 1/2 and 3/2. -
A variable circle c has equation x2 + y2 - 2 (t2 - 3 t + 1) x - 2 (t2 + 2 t) y + t = 0The number t is a parameter. Calculate the locus of the center of the circle.
The center of the circle is C(t2 - 3 t + 1, t2 + 2 t). The center is the intersection point of the linesx = t2 - 3 t + 1 and y = t2 + 2 t
We eliminate t and we find the equation of a parabola.x2 - 2 x y + y2 - 12 x - 13 y + 11 = 0
-
We have, in an orthonormal coordinate system, a circle c through the origin O and with a fixed radius R. Now, the circle c starts to rotate about the origin O. From a fixed point at infinity, we draw tangent lines at the rotating circle. Calculate the locus of all points of tangency.
In most problems the difficulty of calculations depends highly on the choice of the coordinate system.Here we choose the coordinate system such that the given point at infinity is (1,0,0).
The variable center point of c has coordinates (R cos(t), R sin(t)). The number t is the parameter.
The equation of the variable circle is
(x - R cos(t))2 + (y - R sin(t))2 = R2 (1)
The points of tangency are on the tangent chord. It is the polar line of point (1,0,0). It is also the line through the center point of the circle othogonal to the x-axis. The equation of that line isx = R cos(t) (2)
(1) and (2) are the associated curves. We eliminate the parameter t from this two equations.(2) in (1) gives (y - R sin(t))2 = R2 <=> y - R sin(t) = R or y - R sin(t) = -R <=> R sin(t) = y - R or R sin(t) = y + R (3)We have to eliminate t from (2) and (3).(2)2 + (3)2 gives x2 + (y - R)2 = R2 or x2 + (y + R)2 = R2The locus consists of two circels with center points (0,R) and (0,-R) and radius R. -
Calculate the locus of a point P such that the tangent lines from P at the ellipse b2x2 + a2y2 - a2b2 = 0 are orthogonal lines.
The quadratic equation of the tangent lines through point P(x1,y1,1) has equation:(x.Fx' (x1,y1,1) + y.Fy' (x1,y1,1) + Fz' (x1,y1,1))2 - 4 F(x,y,1).F(x1,y1,1) = 0 For the ellipse, this equation becomes: (x.2b2 x1 + y.2a2 y1 -2a2 b2)2 -4(b2 x12 + a2 y12 - a2 b2)(b2 x2 + a2 y2 - a2 b2) = 0 The tangent lines are orthogonal lines if and only if the sum of the coefficients of x2 and y2 is 0. This condition is here 4b4 x12 - 4(b2 x12 + a2 y12 - a2 b2)b2 + 4 a4 y12 -4((b2 x12 + a2 y12 - a2 b2)a2 = 0 <=> -a2 b2 y12 + a2 b4 -a2 b2 x12 + a4 b2 = 0 <=> x12 + y12 = a2 + b2The equation of the locus is the circle x2 + y2 = a2 + b2
Information Provided by Johan.Claeys@ping.be