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Intersection of a 45 degree angle and an ellipse

by jjredfish
Tags: angle, degree, ellipse, intersection
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jjredfish
#1
Apr4-14, 09:18 AM
P: 19
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).
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Mark44
#2
Apr4-14, 10:18 AM
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Quote Quote by jjredfish View Post
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).
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.
jjredfish
#3
Apr4-14, 10:27 AM
P: 19
Quote Quote by Mark44 View Post
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.
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
#4
Apr4-14, 10:32 AM
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P: 21,409
Intersection of a 45 degree angle and an ellipse

Quote Quote by jjredfish View Post
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.
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
#5
Apr4-14, 10:45 AM
P: 19
Quote Quote by Mark44 View Post
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

He went from: x^2 + ((x^2 − 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. combining the line equation with the ellipse equation, using the quadratic equation.

And from that, I got: 1.36x^2 - .8x - .2 = 0
Mark44
#6
Apr4-14, 10:50 AM
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P: 21,409
Quote Quote by jjredfish View Post
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
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.
jjredfish
#7
Apr4-14, 10:58 AM
P: 19
Quote Quote by Mark44 View Post
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.
Thanks again for your help, Mark.

I am coming at this from a very low level of mathematical ability. I can follow along, step-by-step, when I see it, but I don't know how to solve things with the quadratic formula and I don't know how that would get turned into (x,y) that I can graph.
Mark44
#8
Apr4-14, 11:51 AM
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P: 21,409
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.
jjredfish
#9
Apr4-14, 10:13 PM
P: 19
Quote Quote by Mark44 View Post
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.
Got it! Brilliant. Thanks so much, Mark.


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