-
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 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
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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 if
p 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.
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|
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 ...
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|
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
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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)
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|
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 |
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|
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
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|
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)
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|
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)
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|
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
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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 proportional
Since (2,-1-i,4-4i) = (1-i).(1+i,-i,4),
the lines k and l are conjugate imaginary lines.
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|
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)
-
Calculate the components of the curve
x2 + 4 x - 6 = 0
|
x2 + 4 x - 6 = 0
<=>
____ ____
x = V 10 - 2 or x = - V 10 - 2
These 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
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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= 0
|
The 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= 0
|
The 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 = 0
All 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
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|
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 = 0
The 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= 0
The equation of the system of conic sections with
basic points A(1,2); B(2,0); C(-1,1); D(0,3) is
k(-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) = 0
k 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= 0
The 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
-
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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= 0
The system has equation:
k(x + y - 3 z)(x + 2 y - 2 z) + l(2 x + y - 4 z)2= 0
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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) = 0
is 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)2
The 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) = 0
The 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.
-
Calculate the polar line p of P(1,0,1) with respect to the conic section
(x - y) (x + y - 2 z) = 0
This 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 = 0
The 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.
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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 = 0
The coordinates of C are (221,-11,14).
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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.
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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 = 2
The 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 = 0
All 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 form
a 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 = 0
The general equation of a conic section with the
point (0,0,1) as a center-point is
a x2 + 2 b" x y + a' y2 + a" z2= 0
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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 equation
Fx' (x,y,z) - Fy' (x,y,z) = 0
<=>
2 x + 2 y - 5 z - (2 x + 14 y - 17 z) = 0
<=>
z - y = 0
So, 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 = 0
|
The 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 = 0
|
The 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
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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 = 1/3
The directions (1,-2,0) and (1,1/3,0) are conjugated.
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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.
-
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 is
y2 = 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 r
It 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.
-
|
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 is
x = 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 = R2
The locus consists of two circels with center points (0,R) and (0,-R) and
radius R.
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|
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 + b2
The equation of the locus is the circle x2 + y2 = a2 + b2