Problems about Analytic Geometry

• 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 problems

A 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(x_o,y_o), 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 xo

In 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(x_o,y_o) of a parabola, we know that the slope is p/y_o .
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 if

·  
·  p p
·  -- . -- = -1
·  y1 y2
· 
· <=> y1.y2 = - p.p

Point D(x_o,y_o) 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(x_o,y_o) of a parabola, we know that the slope is p/y_o .
The slope of the normal is -y_o/p.
The normal line has equation (y-y_o) = (-y_o/p)(x-x_o).
The coordinates of N are (x_o + p,0).
The coordinates of E are (x_o,0).
The distance |E,N| is p.

About an ellipse

Level 1 problems

Calculate the equation of the normal in a point P(x_o,y_o) of the ellipse

 
 b2 x2 + a2 y2 = a2 b2

We know that the tangent line in a point P(x_o,y_o) is

·  
·  a2 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

Level 2 problems

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

Level 3 problems

Take all chords, with slope m, of an ellipse. Show that all midpoints of these chords are on one line.

Take the ellipse E

·  
·  2 2
·  x y
·  -- + -- = 1
·  2 2
·  a b
· 
· <=>
·  b2 x2 + a2 y2 = a2 b2

and 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 + t

Substitution gives

·  
·  b2 x2 + a2 (m x + t)2 - a2 b2 = 0
· <=>
·  (b2 + m2 a2) x2 + 2 a2 t m x + t2 a2 - b2 a2 = 0

Say 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 is

·  
·  x1 + 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 a2

The 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 gives

·  
·  m y a2 + x b2= 0
· <=>
·  b2 x
·  y = - ----
·  a2 m

All 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 is

·  
·  b tan(t)
·  -------------
·  a(sec(t)- 1)

The slope of DP'is

·  
·  b tan(t)
·  -------------
·  a(sec(t)+ 1)

The product of the slopes of DP and DP' is

·  
·  b2tan2(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

Level 2 problems

A point P(x_o,y_o) is on the hyperbola H with foci F and F'.
Prove that |D,F| = | a - (c/a) x_o |.

Take x_o = a sec(t) and y_o = b tan(t) . Then P(x_o,y_o) 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(x_o,y_o) 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)
·