I would like to find the equation of the following locus.
For a big circle C centered at (0,0), the locus of points that the sum of distances to Y-axis and to C is 1, say in the first quadrant, is a parabola. What is the locus if we replace C by a big ellipse $(\frac xa)^2+(\frac yb)^2=1$?
Hint:
For the circle, we start with $x+r-\sqrt{x^2+y^2}=1$ and get $y^2=2(r-1)x+(r-1)^2$. But for the ellipse E, the distance from (x,y) to E is quite complicate. We might have to deal with the closest point (u,v) on E, i.e. (x,y)(u,v) is perpendicular to E, i.e. $\frac{y-v}{x-u}=\frac{a^2v}{b^2u}$.
Best Answer
The solution is found! The locus (u,v) is very complicate in v. Help from Mathematica.
$ v=\left(\frac{a}{b x}(u-x)+\frac{b}{a}\right)\sqrt{a^2-x^2} $ $ x=\frac{u}{2}+\frac{1}{2} \surd \left(u^2-\frac{2 \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)}{3 \left(a^2-b^2\right)}+\left(2^{1/3} \left(a^4-b^2+2 b^2 u-a^2 u^2\right)^2\right)/\left(3 \left(a^2-b^2\right) t\right)+\frac{1}{3 2^{1/3} \left(a^2-b^2\right)}t\right)+\frac{1}{2} \surd \left(2 u^2-\frac{4 \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)}{3 \left(a^2-b^2\right)}-\left(2^{1/3} \left(a^4-b^2+2 b^2 u-a^2 u^2\right)^2\right)/\left(3 \left(a^2-b^2\right) t\right)-\frac{1}{3 2^{1/3} \left(a^2-b^2\right)}t+\left(-\frac{16 a^4 u}{a^2-b^2}+8 u^3-\frac{8 u \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)}{a^2-b^2}\right)/\left(4 \surd \left(u^2-\frac{2 \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)}{3 \left(a^2-b^2\right)}+\left(2^{1/3} \left(a^4-b^2+2 b^2 u-a^2 u^2\right)^2\right)/\left(3 \left(a^2-b^2\right) t\right)+\frac{1}{3 2^{1/3} \left(a^2-b^2\right)}t\right)\right)\right) $ $ t=\left(108 a^8 \left(a^2-b^2\right) u^2-108 a^4 u^2 \left(a^2 u-b^2 u\right)^2+72 a^4 \left(a^2-b^2\right) u^2 \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)+36 a^4 u \left(a^2 u-b^2 u\right) \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)+2 \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)^3+\surd \left(-4 \left(-12 a^4 \left(a^2-b^2\right) u^2+12 a^4 u \left(a^2 u-b^2 u\right)+\left(-a^4+b^2-2 b^2 u+a^2 u^2\right)^2\right)^3+\left(108 a^8 \left(a^2-b^2\right) u^2-108 a^4 u^2 \left(a^2 u-b^2 u\right)^2+72 a^4 \left(a^2-b^2\right) u^2 \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)+36 a^4 u \left(a^2 u-b^2 u\right) \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)+2 \left(-a^4+b^2-2 b^2 u+a^2 u^2\right)^3\right)^2\right)\right)^{1/3} $