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).
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Intersection of a 45 degree angle and an ellipse
- Thread starter beltran
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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).
Equation of the ellipse:
x^2 + (y/0.6)^2 = 1..........................................(1)
Assuming the line has positive slope = 1 ....................................edited
The equation of the line is:
(y - 0.6) = 1 * (x - 1).................................(2)....................................edited
Now solve for x and y from (1) and (2)
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Equation of the ellipse:
x^2 + (y/0.6)^2 = 1..........................................(1)
Assuming the line has positive slope = 1 ....................................edited
The equation of the line is:
(y - 0.6) = 1 * (x - 1).................................(2)....................................edited
Now solve for x and y from (1) and (2)
Thanks very much for your reply Subhotosh Kahn. Unfortunately, my math is very rusty. I don't remember how to solve this for x and y.
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Hello, beltran!
I hope you aren't too rusty . . .
Subhotosh gave you the equations.
. . Ellipse: .\(\displaystyle \frac{x^2}{1^2} + \frac{y^2}{(\frac{3}{5})^2} \;=\;1 \quad\Rightarrow\quad 9x^2 + 25y^2 \;=\;9\;\;[1]\)
. . Line: .\(\displaystyle y \;=\;x - \frac{2}{5}\;\;[2]\)
Substitute [2] into [1]:
. . \(\displaystyle 9x^2 + 25\left(x-\tfrac{2}{5}\right)^2 \;=\;9 \)
. . \(\displaystyle 9x^2 + 25\left(x^2 - \tfrac{4}{5}x + \tfrac{4}{25}\right) \;=\;9 \)
. . \(\displaystyle 9x^2 + 25x^2 - 20x + 4 \;=\;9\)
. . \(\displaystyle 34x^2 - 20x - 5 \;=\;0\)
Quadratic Formula:
. . \(\displaystyle x \;=\;\frac{20\,\pm\sqrt{20^2 - 4(34)(-5)}}{2(34)} \;=\;\frac{20\,\pm\sqrt{1080}}{68} \;=\;\frac{20\pm6\sqrt{30}}{68}\)
. . \(\displaystyle x \;=\;\frac{10\,\pm\,3\sqrt{30}}{34}\)
\(\displaystyle y \;=\;x - \frac{2}{5} \;=\;\frac{10\,\pm\,3\sqrt{30}}{34} - \frac{2}{5}\)
. . \(\displaystyle y \;=\;\frac{-18\,\pm\,15\sqrt{30}}{170}\)
The point in Quadrant I is: .\(\displaystyle \left(\dfrac{10\,+\,3\sqrt{30}}{34},\;\dfrac{-18\,+\,15\sqrt{30}}{170}\right)\)
I hope you aren't too rusty . . .
Subhotosh gave you the equations.
. . Ellipse: .\(\displaystyle \frac{x^2}{1^2} + \frac{y^2}{(\frac{3}{5})^2} \;=\;1 \quad\Rightarrow\quad 9x^2 + 25y^2 \;=\;9\;\;[1]\)
. . Line: .\(\displaystyle y \;=\;x - \frac{2}{5}\;\;[2]\)
Substitute [2] into [1]:
. . \(\displaystyle 9x^2 + 25\left(x-\tfrac{2}{5}\right)^2 \;=\;9 \)
. . \(\displaystyle 9x^2 + 25\left(x^2 - \tfrac{4}{5}x + \tfrac{4}{25}\right) \;=\;9 \)
. . \(\displaystyle 9x^2 + 25x^2 - 20x + 4 \;=\;9\)
. . \(\displaystyle 34x^2 - 20x - 5 \;=\;0\)
Quadratic Formula:
. . \(\displaystyle x \;=\;\frac{20\,\pm\sqrt{20^2 - 4(34)(-5)}}{2(34)} \;=\;\frac{20\,\pm\sqrt{1080}}{68} \;=\;\frac{20\pm6\sqrt{30}}{68}\)
. . \(\displaystyle x \;=\;\frac{10\,\pm\,3\sqrt{30}}{34}\)
\(\displaystyle y \;=\;x - \frac{2}{5} \;=\;\frac{10\,\pm\,3\sqrt{30}}{34} - \frac{2}{5}\)
. . \(\displaystyle y \;=\;\frac{-18\,\pm\,15\sqrt{30}}{170}\)
The point in Quadrant I is: .\(\displaystyle \left(\dfrac{10\,+\,3\sqrt{30}}{34},\;\dfrac{-18\,+\,15\sqrt{30}}{170}\right)\)