Develop Reference

Draw ellipse passing through 3 points

I have next task - draw ellipe trough 3 points (like in the picture).
.
User can drag theese points that to change ellipse size. Two of points placed on edges of great axis. In my solution I use GDI throug Graphics class.
Now my solution is calculate rectangle corresponding to an ellipse and use
DrawEllipse method and rotate transform (if it's need). But this solution has many mathematical calculation. Maybe there is the simplest approach to solve this task?

Here is how to use the DrawEllipse method from a rotation, the minor axis and two vertices.
First we calculate the Size of the bounding Rectangle:
Given the Points A and B sitting on the short sides of length smallSize we get the long side with a little Pythagoras:
int longSide = (int)(Math.Sqrt((A.Y - B.Y) * (A.Y - B.Y) + (B.X - A.X) * (B.X - A.X)));
So :
Size size = new System.Drawing.Size(longSide, smallSize);
Next we need the rotation angle:
float angle = -(float)(Math.Atan2(A.Y - B.Y, B.X - A.X) * 180f / Math.PI);
And it will make things easier to also get the center Point C:
Point C = new Point((A.X + B.X)/ 2, (A.Y + B.Y)/ 2);
The last thing we want is a routine that draws an ellipse of a given Size, rotated around C at an angle:
void DrawEllipse(Graphics G, Pen pen, Point center, Size size, float angle)
{
int h2 = size.Height / 2;
int w2 = size.Width / 2;
Rectangle rect = new Rectangle( new Point(center.X - w2, center.Y - h2), size );
G.TranslateTransform(center.X, center.Y);
G.RotateTransform(angle);
G.TranslateTransform(-center.X, -center.Y);
G.DrawEllipse(pen, rect);
G.ResetTransform();
}
[![enter image description here][1]][1]
Here is a little testbed that brings it all together:
Point A = new Point(200, 200); // *
Point B = new Point(500, 250);
int smallSize = 50;
void doTheDraw(PictureBox pb)
{
Bitmap bmp = new Bitmap(pb.Width, pb.Height);
float angle = -(float)(Math.Atan2(A.Y - B.Y, B.X - A.X) * 180f / Math.PI);
int longSide = (int)(Math.Sqrt((A.Y - B.Y) * (A.Y - B.Y) + (B.X - A.X) * (B.X - A.X)));
Point C = new Point((A.X + B.X) / 2, (A.Y + B.Y) / 2);
Size size = new System.Drawing.Size((int)longSide, smallSize);
using (Pen pen = new Pen(Color.Orange, 3f))
using (Graphics g = Graphics.FromImage(bmp))
{
// a nice background grid (optional):
DrawGrid(g, 0, 0, 100, 50, 10,
Color.LightSlateGray, Color.DarkGray, Color.Gainsboro);
// show the points we use (optional):
g.FillEllipse(Brushes.Red, A.X - 4, A.Y - 4, 8, 8);
g.FillRectangle(Brushes.Red, B.X - 3, B.Y - 3, 7, 7);
g.FillEllipse(Brushes.Red, C.X - 5, C.Y - 5, 11, 11);
// show the connection line (optional):
g.DrawLine(Pens.Orange, A, B);
// here comes the ellipse:
DrawEllipse(g, pen, C, size, angle);
}
pb.Image = bmp;
}
The grid is a nice helper:
void DrawGrid(Graphics G, int ox, int oy,
int major, int medium, int minor, Color c1, Color c2, Color c3)
{
using (Pen pen1 = new Pen(c1, 1f))
using (Pen pen2 = new Pen(c2, 1f))
using (Pen pen3 = new Pen(c3, 1f))
{
pen2.DashStyle = DashStyle.Dash;
pen3.DashStyle = DashStyle.Dot;
for (int x = ox; x < G.VisibleClipBounds.Width; x += major)
G.DrawLine(pen1, x, 0, x, G.VisibleClipBounds.Height);
for (int y = oy; y < G.VisibleClipBounds.Height; y += major)
G.DrawLine(pen1, 0, y, G.VisibleClipBounds.Width, y);
for (int x = ox; x < G.VisibleClipBounds.Width; x += medium)
G.DrawLine(pen2, x, 0, x, G.VisibleClipBounds.Height);
for (int y = oy; y < G.VisibleClipBounds.Height; y += medium)
G.DrawLine(pen2, 0, y, G.VisibleClipBounds.Width, y);
for (int x = ox; x < G.VisibleClipBounds.Width; x += minor)
G.DrawLine(pen3, x, 0, x, G.VisibleClipBounds.Height);
for (int y = oy; y < G.VisibleClipBounds.Height; y += minor)
G.DrawLine(pen3, 0, y, G.VisibleClipBounds.Width, y);
}
}
Note that I made A, B, smallSide class level variables so I can modify them during my tests, (and I did *)..
As you can see I have added a TrackBar to make the smallside dynamic; for even more fun I have added this MouseClick event:
private void pictureBox1_MouseClick(object sender, MouseEventArgs e)
{
if (e.Button.HasFlag(MouseButtons.Left)) A = e.Location;
else B = e.Location;
doTheDraw(pictureBox1);
}
I hope some of my logic can help you out on what you're trying to achieve. Next question you make try to add more info and be less generic.

Compute the similarity transform (translation and rotation) that brings the first two points to (-a, 0) and (a, 0). [This can be done with a matrix transform or complex numbers.]
Apply the same transform to the third point. Now the equation of the reduced ellipse is
x² / a² + y² / b² = 1.
You simply determine the parameter b by plugging the reduced coordinates of the third point,
b = y / √(1 - x²/a²).
Now you have
the center,
the orientation of the axis,
the length of the major axis,
the length of the minor axis.

Draw an ellipse with a specified "fatness" between 2 points

I have a C# bitmap object, and i am able to draw a line from point A to point B.
I have the 2 points on the edges of the diagram, and I would like to draw an ellipse from A to B. The basic g.DrawEllipse() only draws ellipses either perfectly horizontally or vertically, however I need the ellipse to be kind of diagonal from the one end of the image to the other.
My bitmap: 200 tall by 500 wide
Point A: Column 0, Row 20 (0,20)
Point B: Column 499, Row 60 (499, 60)
Widest Point: 30 - Narrow Radius of the ellipse
Here is what I have so far, the draw ellipse doesnt have the overload I need, so help there please:
using (Graphics g = Graphics.FromImage(bmp))
{
g.DrawLine(pen, new Point(20,0), new Point(499,60));
g.DrawEllipse(pen, 20, 0, someWidth, someHeight);
}

Here is how to use the DrawEllipse method from a rotation, the minor axis and two vertices.
First we calculate the Size of the bounding Rectangle:
Given the Points A and B sitting on the short sides of length smallSize we get the long side with a little Pythagoras:
int longSide = (int)(Math.Sqrt((A.Y - B.Y) * (A.Y - B.Y) + (B.X - A.X) * (B.X - A.X)));
So :
Size size = new System.Drawing.Size(longSide, smallSize);
Next we need the rotation angle:
float angle = -(float)(Math.Atan2(A.Y - B.Y, B.X - A.X) * 180f / Math.PI);
And it will make things easier to also get the center Point C:
Point C = new Point((A.X + B.X)/ 2, (A.Y + B.Y)/ 2);
The last thing we want is a routine that draws an ellipse of a given Size, rotated around C at an angle:
void DrawEllipse(Graphics G, Pen pen, Point center, Size size, float angle)
{
int h2 = size.Height / 2;
int w2 = size.Width / 2;
Rectangle rect = new Rectangle( new Point(center.X - w2, center.Y - h2), size );
G.TranslateTransform(center.X, center.Y);
G.RotateTransform(angle);
G.TranslateTransform(-center.X, -center.Y);
G.DrawEllipse(pen, rect);
G.ResetTransform();
}
Here is a little testbed that brings it all together:
Point A = new Point(200, 200); // *
Point B = new Point(500, 250);
int smallSize = 50;
void doTheDraw(PictureBox pb)
{
Bitmap bmp = new Bitmap(pb.Width, pb.Height);
float angle = -(float)(Math.Atan2(A.Y - B.Y, B.X - A.X) * 180f / Math.PI);
int longSide = (int)(Math.Sqrt((A.Y - B.Y) * (A.Y - B.Y) + (B.X - A.X) * (B.X - A.X)));
Point C = new Point((A.X + B.X) / 2, (A.Y + B.Y) / 2);
Size size = new System.Drawing.Size((int)longSide, smallSize);
using (Pen pen = new Pen(Color.Orange, 3f))
using (Graphics g = Graphics.FromImage(bmp))
{
// a nice background grid (optional):
DrawGrid(g, 0, 0, 100, 50, 10,
Color.LightSlateGray, Color.DarkGray, Color.Gainsboro);
// show the points we use (optional):
g.FillEllipse(Brushes.Red, A.X - 4, A.Y - 4, 8, 8);
g.FillRectangle(Brushes.Red, B.X - 3, B.Y - 3, 7, 7);
g.FillEllipse(Brushes.Red, C.X - 5, C.Y - 5, 11, 11);
// show the connection line (optional):
g.DrawLine(Pens.Orange, A, B);
// here comes the ellipse:
DrawEllipse(g, pen, C, size, angle);
}
pb.Image = bmp;
}
The grid is a nice helper:
void DrawGrid(Graphics G, int ox, int oy,
int major, int medium, int minor, Color c1, Color c2, Color c3)
{
using (Pen pen1 = new Pen(c1, 1f))
using (Pen pen2 = new Pen(c2, 1f))
using (Pen pen3 = new Pen(c3, 1f))
{
pen2.DashStyle = DashStyle.Dash;
pen3.DashStyle = DashStyle.Dot;
for (int x = ox; x < G.VisibleClipBounds.Width; x += major)
G.DrawLine(pen1, x, 0, x, G.VisibleClipBounds.Height);
for (int y = oy; y < G.VisibleClipBounds.Height; y += major)
G.DrawLine(pen1, 0, y, G.VisibleClipBounds.Width, y);
for (int x = ox; x < G.VisibleClipBounds.Width; x += medium)
G.DrawLine(pen2, x, 0, x, G.VisibleClipBounds.Height);
for (int y = oy; y < G.VisibleClipBounds.Height; y += medium)
G.DrawLine(pen2, 0, y, G.VisibleClipBounds.Width, y);
for (int x = ox; x < G.VisibleClipBounds.Width; x += minor)
G.DrawLine(pen3, x, 0, x, G.VisibleClipBounds.Height);
for (int y = oy; y < G.VisibleClipBounds.Height; y += minor)
G.DrawLine(pen3, 0, y, G.VisibleClipBounds.Width, y);
}
}
Note that I made A, B, smallSide class level variables so I can modify them during my tests, (and I did *)..
As you can see I have added a TrackBar to make the smallside dynamic; for even more fun I have added this MouseClick event:
private void pictureBox1_MouseClick(object sender, MouseEventArgs e)
{
if (e.Button.HasFlag(MouseButtons.Left)) A = e.Location;
else B = e.Location;
doTheDraw(pictureBox1);
}
Note that I didn't care for disposing of the old Bitmap; you should, of course..!

If you wish to use Graphics to create a diagonal ellipse, perhaps you can use DrawBezier() method.
Here is some code that does it:
// Draws an ellipse using 2 beziers.
private void DrawEllipse(Graphics g, PointF center, float width, float height, double rotation)
{
// Unrotated ellipse frame
float left = center.X - width / 2;
float right = center.X + width / 2;
float top = center.Y - height / 2;
float bottom = center.Y + height / 2;
PointF p1 = new PointF(left, center.Y);
PointF p2 = new PointF(left, top);
PointF p3 = new PointF(right, top);
PointF p4 = new PointF(right, center.Y);
PointF p5 = new PointF(right, bottom);
PointF p6 = new PointF(left, bottom);
// Draw ellipse with rotated points.
g.DrawBezier(Pens.Black, Rotate(p1, center, rotation), Rotate(p2, center, rotation), Rotate(p3, center, rotation), Rotate(p4, center, rotation));
g.DrawBezier(Pens.Black, Rotate(p4, center, rotation), Rotate(p5, center, rotation), Rotate(p6, center, rotation), Rotate(p1, center, rotation));
}
// Rotating a given point by given angel around a given pivot.
private PointF Rotate(PointF point, PointF pivot, double angle)
{
float x = point.X - pivot.X;
float y = point.Y - pivot.Y;
double a = Math.Atan(y / x);
if (x < 0)
{
a += Math.PI;
}
float size = (float)Math.Sqrt(x * x + y * y);
double newAngel = a + angle;
float newX = ((float)Math.Cos(newAngel) * size);
float newY = ((float)Math.Sin(newAngel) * size);
return pivot + new SizeF(newX, newY);
}
The above code computes the frame of the ellipse (proir to the rotation) at points p1, p2, ..., p6. And then, draws the ellipse as 2 beziers with the ellipse frame rotated points.

Drawing triangles given user input

So I'm trying to make a Program that will draw a triangle given some user input. The variables that the user provides are angleA, angleB, andleC, and the corresponding sides. The code I have set up to find the three points of the angle is as follows.
double angle_A = double.Parse(angleA.Text);
double angle_B = double.Parse(angleB.Text);
double angle_C = double.Parse(angleC.Text);
double side_A = double.Parse(sideA.Text);
double side_B = double.Parse(sideB.Text);
double side_C = double.Parse(sideC.Text);
double triangleHeight = Area * 2 / (double.Parse(sideB.Text));
double height = canvas.Height;
double width = canvas.Width;
int aX, aY, bX, bY, cX, cY;
aY = Convert.ToInt32(canvas.Height - triangleHeight / 2);
if (angle_A <= 90 && angle_C <= 90)
{
aX = Convert.ToInt32((width - side_B) / 2);
}
else if (angle_A > 90)
{
double extraLength = Math.Sqrt(Math.Pow(side_C, 2) - Math.Pow(triangleHeight, 2));
aX = Convert.ToInt32(width - ((width - (side_B + extraLength)) / 2) + side_B);
}
else if (angle_C > 90)
{
double extraLength = Math.Sqrt(Math.Pow(side_A, 2) - Math.Pow(triangleHeight, 2));
aX = Convert.ToInt32((width - side_B + extraLength) / 2);
}
else
{
aX = 0;
MessageBox.Show("ERROR: No such triangle exists", "ERROR:");
}
cX = aX + Convert.ToInt32(side_B);
cY = aY;
bX = Convert.ToInt32(side_A * Math.Cos(Math.PI * angle_C / 180) + cX);
bY = Convert.ToInt32(side_A * Math.Sin(Math.PI * angle_C / 180) - cY);
Point pointA = new Point(aX, aY);
Point pointB = new Point(bX, bY);
Point pointC = new Point(cX, cY);
Point[] points = new Point[3] { pointA, pointB, pointC };
return points;
This returns the three points that the paint method should use to draw the triangle. However, when I insert the values, the triangle it draws looks nothing like the triangle I have described with the user input. Any thoughts on why this is? Thanks in advance.
P.S. The error is not in my code, as it gives me no errors and does not crash. It is strictly a math error that I have not been able to locate.

I imagine the triangle ABC with corners A and C along the base line with A to the left and C to the right, and B somewhere above them. Side A is the side opposite corner A, and so on.
As Damien_the_Unbeliever says, you should only allow input of, say, side B, side C and angle of corner A. Validate that A is not over 180 degrees. Start off with A at the origin, so we know straight away that xA = 0, yA = 0, xC = length of side B, yC=0, xB = side C * cos A, and yB = side C * sin A. I believe this works even if A is over 90 degrees, you do get a negative value for xB but don't worry, continue anyway!
Now all you have to do is centre the triangle on the canvas. I don't understand where you are getting Area from. It makes no sense to calculate the triangle's height from its area. The triangle height is yB, you can calculate the offset you need to centre it vertically as you know, so long as yB <= height. Add the same y offset to all the points.
The horizontal offset is a bit more complicated! If xB is negative, I would add an offset to all the x values to bring xB to 0, this positions your triangle at the left side of the canvas, and its width is given by the new xC. If xB is non-negative, the width is the maximum of xC or xB. Then you can calculate the x offset from the width as you know.
I have had time to do some of the code, for example values; this will draw a triangle but not yet centre it:
int sideB = 100;
int sideC = 143;
int angleA = 28;
double angleARadians = Math.PI * angleA / 180.0;
int[] xs = new int[3];
int[] ys = new int[3];
//1st corner is at the origin
xs[0] = 0; ys[0] = 0;
//Then the third corner is along the x axis from there to the length of side B
xs[2] = sideB; ys[2] = 0;
// The second corner is up a bit above the x axis. x could be negative.
// Note, when you draw it, the y axis extends downwards, so a positive y value will be drawn below the x axis.
xs[1] = (int)Math.Round(sideC * Math.Cos(angleARadians));
ys[1] = (int)Math.Round(sideC * Math.Sin(angleARadians));
//If Bx is negative, move all the points over until it's 0
if (xs[1] < 0)
{
int zeroX = xs[1] * -1;
for (int i = 0; i < 3; i++)
{
xs[i] += zeroX;
}
}
// Now centre the triangle on the canvas.
// Firstly find the width of the triangle. Point B could be to the left of A, or between A and C, or to the right of C.
// So the left most point of the triangle is the minimum of A or B, and the right most point is the maximum of B or C.
int minX = Math.Min(xs[0],xs[1]);
int maxX = Math.Max(xs[2], xs[1]);
//The height of the triangle is yB.
int offsetX = (panCanvas.Width - (maxX - minX)) / 2;
int offsetY = (panCanvas.Height - ys[1]) / 2;
//offset all the points by the same amount, to centre the triangle.
for (int i = 0; i < 3; i++)
{
xs[i] += offsetX;
ys[i] += offsetY;
}

Given the three sides of a triangle a, b, and c the coordinates of the vertices are
P=[0,0]
Q=[a,0]
R=[(a^2+c^2-b^2)/(2*a), sqrt(c^2*(2*(a^2+b^2)-c^2)-(a+b)^2*(a-b)^2)/(4*a^2))]
Example, a=6, b=4 and c=8
P=[0,0]
Q=[6,0]
R=[7,√15]

Calculating the Point3Ds of a Cuboid around a Line

There are two Point3D's (A and B) and I want to calculate the points of a cuboid (a,b,c ... h) surrounding the line between A and B like a hull:
There is one degree of freedom, the angle of the cuboid, because it can rotate around the line AB. I am not sure yet if this is a problem.
I tried to calculate a vector normal to AB, D, and then the cross product of AB ⨯ AD = E. In the code, C is A - B so its the offset parallel to AB.
I normalized these three vectors (C, D and E) and multiplied it with an offset to add / subtract them from A and B. It's not quite working yet.
EDIT: see ja72's code for solution
i also implemented a way of finding a normal vector:
double ax = Vector3D.AngleBetween(E, new Vector3D(1, 0, 0));
double ay = Vector3D.AngleBetween(E, new Vector3D(0, 1, 0));
double az = Vector3D.AngleBetween(E, new Vector3D(0, 0, 1));
ax = Math.Abs(ax - 90);
ay = Math.Abs(ay - 90);
az = Math.Abs(az - 90);
if (ax <= ay & ax <= az)
{
n = Vector3D.CrossProduct(E, new Vector3D(1, 0, 0));
}
else if (az <= ax && az <= ay)
{
n = Vector3D.CrossProduct(E, new Vector3D(0, 0, 1));
}
else
{
n = Vector3D.CrossProduct(E, new Vector3D(0, 1, 0));
}
n = normalize(n);

You need two direction vectors. One is along the line AB given by
Vector3D e = Normalize(B-A)
and one to descibe the "up" direction for the cross section. This can be given, or it can be calculated with the following algorithm (with preference towards +y)
if( e.X != 0 || e.Z != 0 )
{
// choose direction perpendicular to line closest to +y direction
Vector3D n = [-e.X*e.Y, e.X*e.X+e.Z*e.Z, -e.Z*e.Y];
} else {
// if line along +y already then choose +z for up vector
Vector3D n = [ 0, 0, 1];
}
Now you can calculate the 3rd direction to form a coordinate system
Vector3D k = Normalize( Cross(e,n) )
And you assemble the 3×3 rotation matrix that transforms local coordinates to world coordinates with
| k.X n.X e.X |
R = | k.Y n.Y e.Y |
| k.Z n.Z e.Z |
The local coordinate have the direction along the line as +z such that
Point3D a = A + R*[w,h,0]
Point3D b = A + R*[-w,h,0]
Point3D c = A + R*[w,-h,0]
Point3D d = A + R*[-w,-h,0]
Point3D e = B + R*[w,h,0]
Point3D f = B + R*[-w,h,0]
Point3D g = B + R*[w,-h,0]
Point3D h = B + R*[-w,-h,0]
where R*[x,y,z] designates a matrix-vector multiplication, w and h are the width and height of the rectangular cross section, and A, B are the point A and B position vectors. Addition here between vectors is element by element.
I have checked the code in my own code and it works. vec3 is alias for a 3D vector, mat3 is alias for 3×3 matrix.
vec3 u=(B-A).Normalized();
vec3 n = vec3.O;
if(Math.Abs(u.X)<=Math.Abs(u.Y)&&Math.Abs(u.X)<=Math.Abs(u.Z))
{
n=new vec3(u.Y*u.Y+u.Z*u.Z, -u.Y*u.X, -u.Z*u.X);
}
else if(Math.Abs(u.Y)<=Math.Abs(u.X)&&Math.Abs(u.Y)<=Math.Abs(u.Z))
{
n=new vec3(-u.X*u.Y, u.X*u.X+u.Z*u.Z, -u.Z*u.Y);
}
else if(Math.Abs(u.Z)<=Math.Abs(u.X)&&Math.Abs(u.Z)<=Math.Abs(u.Y))
{
n=new vec3(-u.X*u.Z, -u.Y*u.Z, u.X*u.X+u.Y*u.Y);
}
vec3 v=n.Cross(u);
mat3 R=mat3.Combine(v, n, u);
var a=A+R*new vec3(wt, ht, 0);
var b=A+R*new vec3(-wt, ht, 0);
var c=A+R*new vec3(wt, -ht, 0);
var d=A+R*new vec3(-wt, -ht, 0);
var e=B+R*new vec3(wt, ht, 0);
var f=B+R*new vec3(-wt, ht, 0);
var g=B+R*new vec3(wt, -ht, 0);
var h=B+R*new vec3(-wt, -ht, 0);

I can't give you a real piece of code, but I can give you an idea.
Ok, suppose you are already have a proper cuboid. I mean it has right width and height.
Lets locate it on plane xy. After that you need to offset it to center (minus offset vector). The last thing is to rotate it according to your line rotation.
Again:
Create cuboid and locate it on xy plane
Move it according to your offsets
Rotate it according to your line rotation
You may use matrix multiplication to achieve this transformations.

How to find a point where a line intersects an ellipse in 2D (C#)

I need to find a point where a line (its origin is ellipse' center) intersects an ellipse in 2D... I can easily find a point on a circle, because I know an angle F and the circle' radius (R):
x = x0 + R * cosF
y = y0 + R * sinF
However I just can't figure how am I supposed to deal with an ellipse... I know it's dimensions (A & B), but what is the way of finding parameter T?!
x = x0 + A * cosT
y = y0 + B * sinT
From what I understand the parameter T (T angle) is not far from the F angle (approximately +-15 degrees in some cases), but I just can't figure how to calculate it!!!
If there is a kind hearted soul, please help me with this problem...

The standard equation of an ellipse, stationed at 0,0, is:
1 = (x)^2 / (a) + (y)^2 / (b)
Where a is 1/2 the diameter on the horizontal axis, and b is 1/2 the diameter on the vertical axis.
you have a line, assuming an equation:
y = (m)(x - x0) + y0
So, let us plug-and-play!
1 = (x)^2 / (a) + (m(x - x0) + y0)^2 / (b)
1 = x^2 / a + (mx + (y0 - mx0))^2 / b
1 = x^2 / a + (m^2 * x^2 + 2mx*(y0 - mx0) + (y0 - mx0)^2) / b
1 = x^2 / a + (m^2 x^2) / b + (2mx*(y0 - mx0) + (y0^2 - 2y0mx0 + m^2*x0^2)) / b
1 = ((x^2 * b) / (a * b)) + ((m^2 * x^2 * a) / (a * b)) + (2mxy0 - 2m^2xx0)/b + (y0^2 - 2y0mx0 + m^2*x0^2)/b
1 = ((bx^2 + am^2x^2)/(ab)) + (x*(2my0 - 2m^2x0))/b + (y0^2 - 2y0mx0 + m^2*x0^2)/b
0 = x^2*((b + a*m^2)/(ab)) + x*((2my0 - 2m^2x0)/b) + (((y0^2 - 2y0mx0 + m^2*x0^2)/b) - 1)
That last equation follows the form of a standard quadratic equation.
So just use the quadratic formula, with:
((b + a*m^2)/(ab))
((2my0 - 2m^2x0)/b)
and
(((y0^2 - 2y0mx0 + m^2*x0^2)/b) - 1)
to get the X values at the intersections; Then, plug in those values into your original line equation to get the Y values.
Good luck!

Don't do it this way. Instead check the equation that forms an ellipse and that forming a line and solve the set:
The ellipse: (x/a)^2 + (y/b)^2 = 1
Your line: y = cx
You know a, b and c, so finding a solution is going to be easy. You'll find two solutions, because the line crosses the ellipse twice.
EDIT: Note I moved your ellipse's center to (0,0). It makes everything easier. Just add (x0,y0) to the solution.

public Hits<float2> EllipseLineIntersection ( float rx , float ry , float2 p1 , float2 p2 )
{
Hits<float2> hits = default(Hits<float2>);
float2 p3, p4;
Rect rect = default(Rect);
{
rect.xMin = math.min(p1.x,p2.x);
rect.xMax = math.max(p1.x,p2.x);
rect.yMin = math.min(p1.y,p2.y);
rect.yMax = math.max(p1.y,p2.y);
}
float s = ( p2.y - p1.y )/( p2.x - p1.x );
float si = p2.y - ( s * p2.x );
float a = ( ry*ry )+( rx*rx * s*s );
float b = 2f * rx*rx * si * s;
float c = rx*rx * si*si - rx*rx * ry*ry;
float radicand_sqrt = math.sqrt( ( b*b )-( 4f * a * c) );
p3.x = ( -b - radicand_sqrt )/( 2f*a );
p4.x = ( -b + radicand_sqrt )/( 2f*a );
p3.y = s*p3.x + si;
p4.y = s*p4.x + si;
if( rect.Contains(p3) ) hits.Push( p3 );
if( rect.Contains(p4) ) hits.Push( p4 );
return hits;
}
public struct Hits<T>
{
public byte count;
public T point0, point1;
public void Push ( T val )
{
if( count==0 ) { point0 = val; count ++; }
else if( count==1 ) { point1 = val; count ++; }
else print("This structure can only fit 2 values");
}
}

I wrote a C# code for your problem and I hope you can find it helpful. the distance function inside this code calculates euclidean distance between two points in space.
wX denotes horizontal radios of ellipse and wY denotes vertical radios.
private PointF LineIntersectEllipse(PointF A, PointF B, float wX, float wY)
{
double dx = B.X - A.X;
double dy = B.Y - A.Y;
double theta = Math.Atan2(dy, dx);
double r = distance(A, B) - ((wX * wY) / Math.Sqrt(Math.Pow(wY * Math.Cos(theta), 2) + Math.Pow(wX * Math.Sin(theta), 2)));
return PointF((float)(A.X + r * Math.Cos(theta)), (float)(A.Y + r * Math.Sin(theta)));
}

Andrew Łukasik posted a good and useful answer, however it is not using regular C# types. As I wrote in the comments, I converted the code using System.Drawing objects PointF and RectangleF. I found out that if the points given as parameters are aligned as a vertical or horizontal line, then "rect" will have a width or a height equal to 0. Then, rect.Contains(point) will return false even if the point is on this line.
I also modified the "Hits" structure to check if the point pushed is not already existing, which is the case if the line is perfectly tangent, then p3 and p4 will have same coordinates, as the exact tangent point is the only crossing point.
Here is the new code taking care of all the cases :
public static Hits<PointF> EllipseLineIntersection0(float rx, float ry, PointF p1, PointF p2)
{
Hits<PointF> hits = default(Hits<PointF>);
PointF p3 = new PointF();
PointF p4 = new PointF();
var rect = default(RectangleF);
rect.X = Math.Min(p1.X, p2.X);
rect.Width = Math.Max(p1.X, p2.X) - rect.X;
rect.Y = Math.Min(p1.Y, p2.Y);
rect.Height = Math.Max(p1.Y, p2.Y) - rect.Y;
float s = (p2.Y - p1.Y) / (p2.X - p1.X);
float si = p2.Y - (s * p2.X);
float a = (ry * ry) + (rx * rx * s * s);
float b = 2f * rx * rx * si * s;
float c = rx * rx * si * si - rx * rx * ry * ry;
float radicand_sqrt = (float)Math.Sqrt((b * b) - (4f * a * c));
p3.X = (-b - radicand_sqrt) / (2f * a);
p4.X = (-b + radicand_sqrt) / (2f * a);
p3.Y = s * p3.X + si;
p4.Y = s * p4.X + si;
if (rect.Width == 0)
{
if (p3.Y >= rect.Y && p3.Y <= rect.Y + rect.Height) hits.Push(p3);
if (p4.Y >= rect.Y && p4.Y <= rect.Y + rect.Height) hits.Push(p4);
}
else if (rect.Height == 0)
{
if (p3.X >= rect.X && p3.X <= rect.X + rect.Width) hits.Push(p3);
if (p4.X >= rect.X && p4.X <= rect.X + rect.Width) hits.Push(p4);
}
else
{
if (rect.Contains(p3)) hits.Push(p3);
if (rect.Contains(p4)) hits.Push(p4);
}
return hits;
}
public struct Hits<T>
{
public byte Count;
public T P0, P1;
public void Push(T val)
{
if (Count == 0) { P0 = val; Count++; }
else if (Count == 1) { if (!P0.Equals(val)) { P1 = val; Count++; } }
else throw new OverflowException("Structure Hits can only fit 2 values.");
}
}