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If you are looking at the upper right quadrant of an ellipse centered at $(0,0)$, with $a=1$ and $b = 0.6$, and there is a $45$ degree line drawn from $(1, 0.6)$, 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).

Voliar
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beltran
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1 Answers1

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Your ellipse has the following equation: $$ \frac{x^2}{1} + \frac{y^2}{0.6^2} = 1. $$ Your line, which passes through $(1, 0.6)$ with $45$ angle has the following equation: $$ y = x - 0.4. $$ Now you need to solve the system of these two equations. Substituting $y = x-0.4$ to the first equation, we get $$ x^2 + \frac{x^2 - 0.8 x + 0.4^2}{0.6^2} = 1. $$ Therefore, we obtain the following quadratic equation: $$ (1+0.6^2) \, x^2 - 0.8 \,x - (0.6^2-0.4^2) = 0. $$ Hence, $$ 1.36 x^2 - 0.8 x - 0.2 = 0. $$ The roots of this quadratic equation will be (if I didn't make a mistake) $$ x_1 = \frac{0.1 - \sqrt{0.027}}{0.34}, \quad x_2 = \frac{0.1 + \sqrt{0.027}}{0.34}. $$ Thus, the points of intersection of line and ellipse will be $$ \left(\frac{0.1 - \sqrt{0.027}}{0.34}, \frac{0.1 - \sqrt{0.027}}{0.34} - 0.4 \right) \\ \left(\frac{0.1 + \sqrt{0.027}}{0.34}, \frac{0.1 + \sqrt{0.027}}{0.34} - 0.4 \right) $$ The second point lies in the upper right quadrant.

Voliar
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  • Hi Voliar, thank you so much for answering. Does the 0.8x come from 2ab? I can solve down to 1.36x^2-.8x-.2=0, where do I go from there? My math is very rusty. – beltran Apr 04 '14 at 14:58
  • The step I am missing is how do I turn 1.36x^2 - .8x -.2 = 0 into (x,y) coordinates that I can graph? Thanks. – beltran Apr 04 '14 at 16:11
  • @beltran I edited my post, look at it. – Voliar Apr 04 '14 at 16:15
  • Thanks so much Voliar. You are very good at explaining things in a way I can understand. If you don't mind, how do you do the roots, going from 1.36x^2 - .8x -.2 = 0 to the next step? – beltran Apr 04 '14 at 16:49
  • @beltran No problem. You just need to use the so-called Discriminant, see here – Voliar Apr 04 '14 at 20:30