**Solving a System of Conic Equations: An Unorthodox Method**

**Introduction**

Problem: Find the solutions to the following systems of equations:

ax^2 + by^2 = c

rx^2 + sy^2 = t

Assume that:

Only a or b can be negative, while c > 0, and

Only r or s can be negative, while t > 0.

Can use the matrix method to solve for x and y?

[ [a, b], [r, s] ] * [ [ x^2 ], [ y^2 ] ] = [ [ c ], [ t ] ]

[ [ x^2 ], [ y^2 ] ] = [ [a, b], [r, s] ]^(-1) * [ [ c ], [ t ] ]

If x^2 and y^2 are both positive, the conic curves have intersecting points.

**Example 1 – Two Ellipses:**

3x^2 + 5y^2 = 10

4x^2 + y^2 = 6

roots:

x^2 = 1.17647058824, x = ± 1.0846522891

y^2 = 1.29411764706, y = ± 1.1375291799

Intersection Points:

( 1.0846522891, 1.1375291799 ), ( -1.0846522891, 1.1375291799 ),

( 1.0846522891, -1.1375291799 ), (- 1.0846522891, -1.1375291799 )

**Example 2 – An Ellipse and a Hyperbola:**

3x^2 + 5y^2 = 10

4x^2 – y^2 = 6

roots:

x^2 = 1.73913043478, x = ± 1.31876094679

y^2 = 0.956521739126, y = ± 0.978019293849

Intersection Points:

( 1.31876094679, 0.978019293849 ), ( 1.31876094679, -0.978019293849 ),

( -1.31876094679, 0.978019293849 ), ( -1.31876094679, -0.978019293849 )

**Example 3 – An Ellipse and a Hyperbola, part II:**

3x^2 + 5y^2 = 10

-4x^2 + y^2 = 6

roots:

x^2 = -.86956217393

y^2 = 2.51273913043

There are no intersection points.

The program SYSCONIC (HP Prime) solves the system. Output: a 6 x 2 matrix:

[ [ x, 0 ]

[ y, 0 ]

[ a*x^2 + b*y^2, c ]

[ r*x^2 + s*y^2, t ]

[ a*(-x)^2 + b*(-y)^2, c ]

[ r*(-x)^2 + s*(-y)^2, t ] ]

The last four rows are to check solutions.

EXPORT SYSCONIC()

BEGIN

HComplex:=11; // complex

// 2019-07-15 EWS

// System conic equations

LOCAL a,b,c,r,s,t;

LOCAL m0,m1,m2,m3,x,y;

INPUT({a,b,c,r,s,t},

“ax^2+by^2=c rx^2+sy^2=t”,

{“a: “,”b: “,”c: “,

“r: “,”s: “,”t: “});

m0:=[[a,b],[r,s]];

m1:=[[c],[t]];

m2:=m0^(−1)*m1;

x:=m2(1,1);

x:=√x;

y:=m2(2,1);

y:=√y;

MSGBOX(x);

MSGBOX(y);

// results and check

m3:=MAKEMAT(0,6,2);

m3(1,1):=x;

m3(2,1):=y;

m3(3,1):=a*x^2+b*y^2;

m3(3,2):=c;

m3(4,1):=r*x^2+s*y^2;

m3(4,2):=t;

m3(5,1):=a*(−x)^2+b*(−y)^2;

m3(5,2):=c;

m3(6,1):=r*(−x)^2+s*(-y)^2;

m3(6,2):=t;

END;

**Remarks**

It is a wise idea to check the answers to make sure that they are correct, this is kind of an unorthodox approach.

If we tried the matrix method on a system of equations like:

x^2 + 4x – y = 2

-3x^2 + x – 2y = 5

the method will not work. Each term must have a different variable.

Eddie

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