Answer to Question #224010 in Analytic Geometry for Bless

Question

A point moves in the Oxy−plane so that the sum of its distances from the points S(2, 2) and S′(6, 2) is 16. Find the equation of the curve traced by the point.

Answer 1

$This \space is \space the \space classic \space problem \space to \space derive \space the \space equation \space of \space an \space ellipse \space \\ with \space foci \space at \space (2, \space 2) \space and \space (6, \space 2).\\ Call \space a \space point \space on \space the \space curve \space (x, \space y). \space Then \space the \space distance \space formula \space says:\\ \displaystyle \space \sqrt{(x \space - \space 2)^2 \space + \space (y \space - \space 2)^2} \space + \space \sqrt{(x \space - \space 6)^2 \space + \space (y \space - \space 2)^2} \space = \space 16\\ Get \space rid \space of \space the \space radicals:\\ \displaystyle \space \sqrt{x^2 \space + \space y^2 \space - \space 4x \space - \space 4y \space + \space 8} \space + \space \sqrt{x^2 \space + \space y^2 \space - \space 12x \space - \space 4y \space + \space 40} \space = \space 16\\ \displaystyle \space \sqrt{x^2 \space + \space y^2 \space - \space 4x \space - \space 4y \space + \space 8} \space = \space 16 \space - \space \sqrt{x^2 \space + \space y^2 \space - \space 12x \space - \space 4y \space + \space 40}\\ \displaystyle \space x^2 \space + \space y^2 \space - \space 4x \space - \space 4y \space + \space 8 \space = \space (16 \space - \space \sqrt{x^2 \space + \space y^2 \space - \space 12x \space - \space 4y \space + \space 40})^2\\ \displaystyle \space x^2 \space + \space y^2 \space - \space 4x \space - \space 4y \space + \space 8 \space = \space 256 \space - \space \displaystyle \space 32 \space \sqrt{x^2 \space + \space y^2 \space - \space 12x \space - \space 4y \space + \space 40} \space + \space (x^2 \space + \space y^2 \space - \space 12x \space - \space 4y \space + \space 40)\\ \displaystyle \space 32 \space \sqrt{x^2 \space + \space y^2 \space - \space 12x \space - \space 4y \space + \space 40} \space = \space 288 \space - \space 8x \space = \space 8(36 \space - \space x)\\ \displaystyle \space 4 \space \sqrt{x^2 \space + \space y^2 \space - \space 12x \space - \space 4y \space + \space 40} \space = \space 36 \space - \space x\\ \displaystyle \space 16 \space (x^2 \space + \space y^2 \space - \space 12x \space - \space 4y \space + \space 40) \space = \space (36 \space - \space x)^2\\ \displaystyle \space 16x^2 \space + \space 16y^2 \space - \space 192x \space - \space 64y \space + \space 640 \space = \space 1296 \space - \space 72x \space + \space x^2\\ \displaystyle \space 15x^2 \space - \space 120x \space + \space 16y^2 \space - \space 64y \space = \space 656\\ Now \space complete \space the \space square \space on \space x \space and \space y:\\ \displaystyle \space 15(x^2 \space - \space 8x) \space + \space 16(y^2 \space - \space 4y) \space = \space 656\\ \displaystyle \space 15(x^2 \space - \space 8x \space + \space 16) \space + \space 16(y^2 \space - \space 4y \space + \space 4) \space = \space 656 \space + \space 15 \space \cdot \space 16 \space + \space 16 \space \cdot \space 4\\ \displaystyle \space 15(x \space - \space 4)^2 \space + \space 16(y \space - \space 2)^2 \space = \space 960\\ Divide \space both \space sides \space by \space 960:\\ \displaystyle \space \frac{(x \space - \space 4)^2}{64} \space + \space \frac{(y \space - \space 2)^2}{60} \space = \space 1\\ hence \space this \space represents \space the \space equation \space of \space an \space ellipse. \space \\$