If the coordinates of a point lies on ellipse 9x^2+16y^2=144 be (2,3root3/2), find the ecentric angle of that point
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The focus of the ellipse[math]\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)=1[/math]is at[math]f={\sqrt {a^{2}-b^{2}}}[/math]and the length of the latus rectum is[math]2\frac{b^2}{a}[/math]. This gives the coordinates an the extremity of the latus rectum as[math]P=\left(\sqrt {a^{2}-b^{2}}, \frac{b^2}{a}\right)[/math]Differentiating wrt x, y gives a not unit length as normal[math]\left(\frac{2x}{a^2},\frac{2y}{b^2}\right)[/math], apply this at the latus rectum gives[math]N=\left(2\frac{\sqrt {a^{2}-b^{2}}}{a^2}, 2\frac{b^2}{a b^2}\right)[/math], and the normal line is then[math]P+\lambda N[/math]. We want this to passthrough the point[math](0,-b)[/math]. Doing the y-coordinate[math]-b= \frac{b^2}{a} +\lambda \frac{2}{a}[/math],[math]-ab= b^2 + 2\lambda[/math]so[math]\lambda=-\frac{ab+b^2}{2}[/math]. For the x-coord[math]0=\sqrt {a^{2}-b^{2}} - \lambda 2\frac{\sqrt {a^{2}-b^{2}}}{a^2}[/math]multiply by a squared and divide by the square root gives [math]\lambda= -a^2[/math]. Setting both to the same value we find[math]a^2-ab-b^2=0[/math]. Let[math]\mu=a/b[/math]we have a quadratic[math]\mu^2-\mu-1[/math]with solutions[math]\mu=\frac{1\pm\sqrt{5}}{2}=\tau[/math]the golden ratio. The eccentricity is[math]e=\sqrt{1-\left(\frac{b}{a}\right)^2}=\sqrt{1-\frac{1}{\tau^2}}[/math]. This simplifies to[math]e=1/\sqrt{\tau}[/math].
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