Some Problems related to linear equation and elliptic equation

My needs are shown in the picture is.

**1. Calculate the equation after a straight line and an ellipse are simultaneous (eliminate x)

1. Calculating the discriminant of simultaneous equations
2. Calculate Veda of simultaneous equations (x1+x2, x1 x2, y1+y2, y1 y2, x1 x2+y1 y2, x1 y2+x2 y1)
3. Calculate the chord AB length where the ellipse intersects the line**

>> syms a b c x y x0 y0 x1 y1 x2 y2 k m t n

>> eqn1=x^2/a^2+y^2/b^2==1;

>> eqn2=y==k*x+m;

>> eqns = [eqn1 eqn2];

>> [(x1,y1) (x2,y2)]=solve([eqn1 eqn2,(x,y)]

[浩思]

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr = collect(eliminate(eqns,y))
expr1=eliminate(eqns,y);
p = coeffs(expr1(1),x);
discriminant=simplify((p(2)^2-4*p(1)*p(3)))
[x,y] = solve(eqns,[x,y]);
P1 = [x(1),y(1)];
P2 = [x(2),y(2)];
x1plusx2=simplify(x(1)+x(2))
x1x2=simplify(x(1)*x(2))
y1plusy2=simplify(y(1)+y(2))
y1y2=simplify(y(1)*y(2))
chord_length = simplify(sqrt((x(1)-x(2))^2+(y(1)-y(2))^2))

[浩思]

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr = collect(eliminate(eqns,y))

expr = 

expr1=eliminate(eqns,y);
p = coeffs(expr1(1),x);
discriminant=simplify((p(2)^2-4*p(1)*p(3)))

discriminant = 

[x,y] = solve(eqns,[x,y]);
P1 = [x(1),y(1)];
P2 = [x(2),y(2)];
x1plusx2=simplify(x(1)+x(2))

x1plusx2 = 

x1x2=simplify(x(1)*x(2))

x1x2 = 

y1plusy2=simplify(y(1)+y(2))

y1plusy2 = 

y1y2=simplify(y(1)*y(2))

y1y2 = 

chord_length = simplify(sqrt((x(1)-x(2))^2+(y(1)-y(2))^2))

chord_length = 

Answer 1

[Torsten]

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
[x,y] = solve(eqns,[x,y])

x = 

y = 

P1 = [x(1),y(1)]

P1 = 

P2 = [x(2),y(2)]

P2 = 

[浩思]

1. Why is my answer one-dimensional rather than two-dimensional?

2. How to assign the two groups of roots of the equation group to x1, x2, y1, and y2 respectively, so as to facilitate the following calculations, such as calculating the values of x1x2, x1+x2

[浩思]

In other words, the result I want is strictly the result shown in the picture above

[Torsten]

You have two solutions for x and two solutions for y.

These are your x1,x2,y1,y2 in case that the discriminant is > 0.

I don't know how you can access them in octave ; in MATLAB you could set

P1 = [x(1),y(1)];
P2 = [x(2),y(2)];

[浩思]

These Vedas are obtained on the premise that the discriminant is greater than 0.

I think the first step is to get the quadratic equation of one variable after the combination of elliptic equation and linear equation. How can I get it? Eliminate y

[浩思]

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
[x,y] = solve(eqns,[x,y]);
P1 = [x(1),y(1)];
P2 = [x(2),y(2)];
x1plusx2=simplify(x(1)+x(2))
x1x2=simplify(x(1)*x(2))
y1plus=simplify(y(1)+y(2))
y1y2=simplify(y(1)*y(2))

[浩思]

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
[x,y] = solve(eqns,[x,y]);
P1 = [x(1),y(1)];
P2 = [x(2),y(2)];
x1plusx2=simplify(x(1)+x(2))

x1plusx2 = 

x1x2=simplify(x(1)*x(2))

x1x2 = 

y1plus=simplify(y(1)+y(2))

y1plus = 

y1y2=simplify(y(1)*y(2))

y1y2 = 

[浩思]

>> syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
[x,y] = solve(eqns,[x,y]);
P1 = [x(1),y(1)];
P2 = [x(2),y(2)];
x1plusx2=simplify(x(1)+x(2))
x1x2=simplify(x(1)*x(2))
y1plusy2=simplify(y(1)+y(2))
y1y2=simplify(y(1)*y(2))

Answer 2

[浩思]

The quadratic equation of one variable after the combination of elliptic equation and linear equation eliminates y. Why there is an error

[浩思]

>> eqn1 = x^2/a^2+y^2/b^2==1;

Invalid use of operator.

eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr=eliminate(eqns,y)

[Torsten]

I don't understand what you are trying to do.

You already have the solutions for the intersection of the line and the ellipse. No need to start anew.

syms a b x y k m eta1 eta2
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
[x,y] = solve(eqns,[x,y]);
eta1 = (k*x(1)+m);
eta2 = (k*x(2)+m);
chord_length = simplify(sqrt((x(1)-x(2))^2+(eta1-eta2)^2))

chord_length = 

[浩思]

This is a math assignment. The requirements are as follows. Score step by step. To solve equations and discriminants simultaneously

[浩思]

>> syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
[x,y] = solve(eqns,[x,y]);
P1 = [x(1),y(1)];
P2 = [x(2),y(2)];
x1plusx2=simplify(x(1)+x(2))
x1x2=simplify(x(1)*x(2))
y1plusy2=simplify(y(1)+y(2))
y1y2=simplify(y(1)*y(2))
chord_length = simplify(sqrt((x(1)-x(2))^2+(y(1)-y(2))^2))

[浩思]

 syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
[x,y] = solve(eqns,[x,y]);
P1 = [x(1),y(1)];
P2 = [x(2),y(2)];
x1plusx2=simplify(x(1)+x(2))

x1plusx2 = 

x1x2=simplify(x(1)*x(2))

x1x2 = 

y1plusy2=simplify(y(1)+y(2))

y1plusy2 = 

y1y2=simplify(y(1)*y(2))

y1y2 = 

chord_length = simplify(sqrt((x(1)-x(2))^2+(y(1)-y(2))^2))

chord_length = 

[Torsten]

So any questions left ?

[浩思]

It is required to write the unitary quadratic equation obtained by eliminating y from elliptic equation and linear equation, and write its discriminant to reach the complete goal. thank you!

-a^2 b^2 + a^2 m^2 + 2 a^2 k m x + (b^2 + a^2 k^2) x^2==0

4 a^2 b^2 (b^2 + a^2 k^2 - m^2)

[Torsten]

@浩思 comment moved here:

How to get such results:

-a^2 b^2 + a^2 m^2 + 2 a^2 k m x + (b^2 + a^2 k^2) x^2==0

And calculate the discriminant:

4 a^2 b^2 (b^2 + a^2 k^2 - m^2)

[浩思]

syms x y a b m k
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr=eliminate(eqns,y)

[浩思]

syms x y a b m k
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr=eliminate(eqns,y)

expr = 

[浩思]

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr = eliminate(eqns,y)
[x,y] = solve(eqns,[x,y]);
P1 = [x(1),y(1)];
P2 = [x(2),y(2)];
x1plusx2=simplify(x(1)+x(2))
x1x2=simplify(x(1)*x(2))
y1plusy2=simplify(y(1)+y(2))
y1y2=simplify(y(1)*y(2))
chord_length = simplify(sqrt((x(1)-x(2))^2+(y(1)-y(2))^2))

[浩思]

Next, solve the problem of discriminant

[浩思]

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr = collect(eliminate(eqns,y))
[x,y] = solve(eqns,[x,y]);
P1 = [x(1),y(1)];
P2 = [x(2),y(2)];
x1plusx2=simplify(x(1)+x(2))
x1x2=simplify(x(1)*x(2))
y1plusy2=simplify(y(1)+y(2))
y1y2=simplify(y(1)*y(2))
chord_length = simplify(sqrt((x(1)-x(2))^2+(y(1)-y(2))^2))

[浩思]

syms a x b c
coeffs(a*x^2 + b*x + c, x)

ans = 

[浩思]

The coefficients of the quadratic equation with one variable are extracted. What code functions can be used to find the discriminant?

[浩思]

syms a x b c
A=coeffs(a*x^2 + b*x + c, x)

A = 

discriminant=A(2)^2-4*A(1)*A(3)

discriminant = 

[浩思]

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr = collect(eliminate(eqns,y))

expr = 

A=coeffs(expr, x);

Error using symengine
Polynomial expression expected.

Error in sym/coeffs (line 59)
cSym = mupadmex('symobj::coeffs',p.s, args{:});

discriminant=A(2)^2-4*A(1)*A(3)
[x,y] = solve(eqns,[x,y]);
P1 = [x(1),y(1)];
P2 = [x(2),y(2)];
x1plusx2=simplify(x(1)+x(2))
x1x2=simplify(x(1)*x(2))
y1plusy2=simplify(y(1)+y(2))
y1y2=simplify(y(1)*y(2))
chord_length = simplify(sqrt((x(1)-x(2))^2+(y(1)-y(2))^2))

[浩思]

Single code runs correctly. Why is it not put together correctly?

[Torsten]

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
p = collect(expand(subs(eqn1,y,solve(eqn2,y))),x);
p = coeffs(lhs(p)-rhs(p),x);
discriminant=simplify(p(2)^2-4*p(1)*p(3))

discriminant = 

[浩思]

The result of this discriminant is incorrect

The correct discriminant is:

4 a^2 b^2 (b^2 + a^2 k^2 - m^2)

[Torsten]

Then multiply by a^2*b^2:

p = coeffs(lhs(p)-rhs(p),x)*a^2*b^2;

instead of

p = coeffs(lhs(p)-rhs(p),x);

It's all just a question of scaling of the polynomial.

What matters is

a^2*k^2 + b^2 - m^2 >= 0

I'd call this expression the "discriminant".

[浩思]

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
p = collect(expand(subs(eqn1,y,solve(eqn2,y))),x);
p = coeffs(lhs(p)-rhs(p),x);
discriminant=simplify((p(2)^2-4*p(1)*p(3))*a^4*b^4)

discriminant = 

[浩思]

Is there a simpler algorithm?

Why can't we directly extract the coefficients of quadratic term, primary term and constant term from the quadratic equation of one variable obtained after simultaneous determination?

-a^2 b^2 + a^2 m^2 + 2 a^2 k m x + (b^2 + a^2 k^2) x^2

[浩思]

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr = collect(eliminate(eqns,y))
p = collect(expand(subs(eqn1,y,solve(eqn2,y))),x);
p = coeffs(lhs(p)-rhs(p),x);
discriminant=simplify((p(2)^2-4*p(1)*p(3))*a^4*b^4)
[x,y] = solve(eqns,[x,y]);
P1 = [x(1),y(1)];
P2 = [x(2),y(2)];
x1plusx2=simplify(x(1)+x(2))
x1x2=simplify(x(1)*x(2))
y1plusy2=simplify(y(1)+y(2))
y1y2=simplify(y(1)*y(2))
chord_length = simplify(sqrt((x(1)-x(2))^2+(y(1)-y(2))^2))

[浩思]

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr = collect(eliminate(eqns,y))

expr = 

p = collect(expand(subs(eqn1,y,solve(eqn2,y))),x);
p = coeffs(lhs(p)-rhs(p),x);
discriminant=simplify((p(2)^2-4*p(1)*p(3))*a^4*b^4)

discriminant = 

[x,y] = solve(eqns,[x,y]);
P1 = [x(1),y(1)];
P2 = [x(2),y(2)];
x1plusx2=simplify(x(1)+x(2))

x1plusx2 = 

x1x2=simplify(x(1)*x(2))

x1x2 = 

y1plusy2=simplify(y(1)+y(2))

y1plusy2 = 

y1y2=simplify(y(1)*y(2))

y1y2 = 

chord_length = simplify(sqrt((x(1)-x(2))^2+(y(1)-y(2))^2))

chord_length = 

[浩思]

Is there a simpler algorithm?

Why can't we directly extract the coefficients of quadratic term, primary term and constant term from the quadratic equation of one variable obtained after simultaneous determination?

-a^2 b^2 + a^2 m^2 + 2 a^2 k m x + (b^2 + a^2 k^2) x^2

[Torsten]

We can:

syms a b m k x
p = coeffs(-a^2*b^2 + a^2*m^2 + 2*a^2*k*m*x + (b^2 + a^2*k^2)*x^2,x)

p = 

discriminant=simplify((p(2)^2-4*p(1)*p(3)))

discriminant = 

But how do you arrive at this term ?

[浩思]

Why is an error reported when running this code?

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr = collect(eliminate(eqns,y))
A=coeffs(expr, x);

[Torsten]

It's pure luck that expr is not

[1 - x^2/a^2 - (k*x+m)^2/b^2]

but

[a^2*b^2 - a^2*k^2*x^2 - 2*a^2*k*m*x - a^2*m^2 - b^2*x^2]

So nothing will work automatically with symbolic computations - all depends on how MATLAB (or octave in your case) chooses to simplify things.

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr = eliminate(eqns,y)

expr = 

p = coeffs(expr(1),x)

p = 

discriminant=simplify((p(2)^2-4*p(1)*p(3)))

discriminant = 

[浩思]

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr = collect(eliminate(eqns,y))

expr = 

expr1=eliminate(eqns,y);
p = coeffs(expr1(1),x);
discriminant=simplify((p(2)^2-4*p(1)*p(3)))

discriminant = 

[x,y] = solve(eqns,[x,y]);
P1 = [x(1),y(1)];
P2 = [x(2),y(2)];
x1plusx2=simplify(x(1)+x(2))

x1plusx2 = 

x1x2=simplify(x(1)*x(2))

x1x2 = 

y1plusy2=simplify(y(1)+y(2))

y1plusy2 = 

y1y2=simplify(y(1)*y(2))

y1y2 = 

chord_length = simplify(sqrt((x(1)-x(2))^2+(y(1)-y(2))^2))

chord_length = 

[浩思]

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr = collect(eliminate(eqns,y))
expr1=eliminate(eqns,y);
p = coeffs(expr1(1),x);
discriminant=simplify((p(2)^2-4*p(1)*p(3)))
[x,y] = solve(eqns,[x,y]);
x1plusx2=simplify(x(1)+x(2))
x1x2=simplify(x(1)*x(2))
y1plusy2=simplify(y(1)+y(2))
y1y2=simplify(y(1)*y(2))
chord_length = simplify(sqrt((x(1)-x(2))^2+(y(1)-y(2))^2))

[浩思]

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr = collect(eliminate(eqns,y))
expr1=eliminate(eqns,y);
p = coeffs(expr1(1),x);
discriminant=simplify((p(2)^2-4*p(1)*p(3)))
[x,y] = solve(eqns,[x,y]);
P1 = [x(1),y(1)];
P2 = [x(2),y(2)];
x1plusx2=simplify(x(1)+x(2))
x1x2=simplify(x(1)*x(2))
y1plusy2=simplify(y(1)+y(2))
y1y2=simplify(y(1)*y(2))
chord_length = simplify(sqrt((x(1)-x(2))^2+(y(1)-y(2))^2))

[浩思]

In this code, what does the 1 in parentheses mean?

expr(1)

p = coeffs(expr(1),x)

[浩思]

Is it the form of quadratic equation with one variable arranged by the ascending power of x?

[Torsten]

No, expression is a vector with 1 element (as you can see by the brackets around it).

expr(1) accesses its first (and only) element.

To be honest: I don't know why this is necessary.

[浩思]

So, if you solve an equation with two roots. Does the software default that x (1) and x (2) are its two roots? Do you want to use x (1) and x (2) to call these two roots without setting them?

[Torsten]

The software writes the solution in a variable that you define as output variable (in the below code: x). If the equation has several solutions, the output variable will be a vector with as many elements as there are solutions. You can access the k'th solution as x(k).

syms x
eq = x^2 == 2;
x = solve(eq,x)

x = 

size(x)

ans = 1×2

2 1

x(1)

ans = 

x(2)

ans = 

[浩思]

I see. This involves the computing essence of the software

[浩思]

Why would an error occur if the 1 in the original code bracket was changed to 2. How can I know how many elements are contained in such parameters (expr)?

The 1 in the brackets of expr has not been understood before

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr = collect(eliminate(eqns,y))

expr = 

expr1=eliminate(eqns,y);
p = coeffs(expr1(2),x)

Index exceeds the number of array elements. Index must not exceed 1.

Error in indexing (line 1079)
R_tilde = builtin('subsref',L_tilde,Idx);

[浩思]

Running expr (1), why can we extract the coefficients of the quadratic term, the coefficient of the primary term and the coefficient of the constant term of the equation after merging the similar terms according to the x term. If expr contains only one element

[Torsten]

a = [1 2 3 4]

a = 1×4

1 2 3 4

This is a row vector with 4 elements. You can get the number of elements with the "size" command:

number_of_elements_of_a = size(a,2)

number_of_elements_of_a = 4

You can access the i-th element of the vector by referencing a(i):

third_element_of_a = a(3)

third_element_of_a = 3

If a vector has only four elements and you try to access the fifth (which does not exist), the error message from above will pop up:

fifth_element_of_a = a(5)

Index exceeds the number of array elements. Index must not exceed 4.

[浩思]

This command coeffs has the function of combining similar items to find coefficients?

[Torsten]

Yes, for polynomials. And you can specify the variable according to which the coefficients are collected.

But you can read all this online in the MATLAB documentation with many examples.

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr = eliminate(eqns,y)

expr = 

coeffs(expr(1),x)

ans = 

[Torsten]

@浩思 comment moved here:

The vector expr only contains one element. Why do you use expr (1), but use expr will report an error?

Does the software not regard the default variable as a vector with only one element?

Thank you very much for your help. I will carefully read the software documentation.

syms a b x y k m 
eqn1 = x^2/a^2+y^2/b^2==1;
eqn2 = y==k*x+m;
eqns = [eqn1 eqn2];
expr = eliminate(eqns,y)

expr = 

coeffs(expr,x)

Error using symengine
Polynomial expression expected.

Error in sym/coeffs (line 59)
cSym = mupadmex('symobj::coeffs',p.s, args{:});

[Torsten]

"coeffs" works on polynomials.

"expr" is a vector. The first element of this vector is a polynomial.

Do you see the brackets [ ] around a^2b^2 - a^2k^2x^2 - 2a^2kmx - a^2m^2 - b^2x^2 ?

They indicate "expr" is a vector (of polynomial expressions), not the polynomial expression itself. Taking the first element of this vector gives the polynomial expression on which you can apply "coeffs".

But I already explained this in a previous comment:

No, expression is a vector with 1 element (as you can see by the brackets around it).

expr(1) accesses its first (and only) element.

To be honest: I don't know why this is necessary.

And please don't add an answer if you want to make a comment.