# Intersection of a 45 degree angle and an ellipse

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).

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.