Do you know how to find the equation of the ellipse? Since you know the major and minor semiaxes (a = 1, b = .6) it's a simple matter write the ellipse's equation.jjredfish said:If you are looking at the upper right quadrant of an ellipse centered at (0,0), with a=1 and b=.6, and there is a 45 degree line drawn from (1,.6), how would I find the (x,y) coordinate where the line crosses the ellipse? (I have been out of school for a long time, this is not homework).
Mark44 said:Do you know how to find the equation of the ellipse? Since you know the major and minor semiaxes (a = 1, b = .6) it's a simple matter write the ellipse's equation.
The line has a slope of 1, and goes through the point (1, .6), so it should be fairly easy to write the line's equation.
Once you have equations for the ellipse and the line, solve the system of two equations for the point of intersection.
Your line equation looks OK, but not your equation for the ellipse.jjredfish said:Hi Mark44, thanks for your reply.
Unfortunately, my math is extremely rusty/non-existent. With someone else's help, I have gotten to y=x-.4, and 1.36x^2-.8x-.2=0 but I don't know how to turn those into (x,y) that I can graph.
Mark44 said:Your line equation looks OK, but not your equation for the ellipse.
For an ellipse whose center is at (0, 0), with vertices at (a, 0) and (0, b), the equation is
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
I don't know how you got 1.36x^2-.8x-.2=0.
BTW, put some spaces in your equations to make them easier to read, like so:
1.36x^2 - .8x -.2 = 0
jjredfish said:He went from: x2 + ((x2 − 0.8x + 0.4^2) / 0.6^2) = 1.
To: (1 + 0.6^2) x^2 − 0.8x − (0.6^2 − 0.4^2) = 0.
And from that, I got: 1.36x^2 - .8x - .2 = 0
Mark44 said:The ellipse equation is .36x2 + y2 = .36
The line equation is y = x - .4
Replace y in the first equation with x - .4, and you'll get an equation that is equivalent to the first one you show above.
This gives you a quadratic equation, which can be solved by the use of the Quadratic Formula.
Mark44 said:Quadratic formula: http://en.wikipedia.org/wiki/Quadratic_formula
Your quadratic equation is 1.36x2 - .8x - .2 = 0. In the quadratic formula, a = 1.36, b = -.8, and c = -.2. You should get two values for x - one positive and one negative. Since you're looking at the intersection of the line and ellipse in the first quadrant, you want the positive x value.
Substitute the value you find in the equation of your line to get the y value. That (x, y) point will be on both the ellipse and the line.
The intersection point of a 45 degree angle and an ellipse is where the line of the angle touches the curve of the ellipse.
To find the intersection point, you can use the equation of the ellipse and the slope of the 45 degree angle. Set the equation of the ellipse equal to the equation of the line and solve for the x and y values.
Yes, there can be multiple intersection points between a 45 degree angle and an ellipse. This will depend on the size and orientation of the ellipse, as well as the position of the angle relative to the ellipse.
The eccentricity of an ellipse, which determines its shape, can affect the position of the intersection point with a 45 degree angle. A higher eccentricity means the ellipse is more elongated and the intersection point may be farther away from the center of the ellipse.
Yes, there are a few special cases for the intersection of a 45 degree angle and an ellipse. If the ellipse is a circle, the intersection point will be at the same distance from the center as the radius of the circle. If the 45 degree angle is parallel to the major or minor axis of the ellipse, there will be no intersection point. And if the angle is perpendicular to the major or minor axis, the intersection point will be at the center of the ellipse.