Find the points on $$ 4x^2+ 9y^2= 36 $$ closest and farthest from $P(1,1)$. I some how ended up with a quartic equation and it has complex roots. I don't know what went wrong.

  • $\begingroup$ How did you get that quartic equation, and what equation is it? $\endgroup$ – Arthur Nov 14 '16 at 16:46
  • $\begingroup$ 46656L^4 - 33696L^3 + 8208L^2 - 792L + 23 = 0 $\endgroup$ – Math Nov 14 '16 at 16:50
  • $\begingroup$ Before you even start calculating, first draw the problem. I think it might be very helpful to understand what you're trying to do. $\endgroup$ – hkr Nov 14 '16 at 16:50
  • $\begingroup$ What exactly is $P(1,1)$? $\endgroup$ – Math1000 Nov 14 '16 at 17:23
  • $\begingroup$ You can parametrize the curve as $x=3\cos t,\ y=2 \sin t.$ The derivative of the squared distance from $(1,1)$ to that, if set to zero, is not really simple, but it's a start. $\endgroup$ – coffeemath Nov 14 '16 at 17:24

The function you are minimizing/maximizing here is the distance function given by $$d(x,y) = \sqrt{(x-1)^2 + (y-1)^2}$$. WLOG, one could minimizing/maximizing the function $d^2=D$ instead since then the algebra is simpler. Define $g(x,y) = 4x^2 + 9y^2 - 36$ and let $\lambda$ be the Lagrange multiplier. Using method of Lagrange multiplier, you need to solve the following system of equations together with the constraint $g(x,y)=0$. \begin{alignat*}{3} 2(x-1) & = D_x && = \lambda g_x && = 8x.\\ 2(y-1) & = D_y && = \lambda g_y && = 18y.\\ & g(x,y) && = 0. \end{alignat*}

  • $\begingroup$ I didn't do the algebra but I trust you can do it yourself. Let me know if anything is unclear thou. $\endgroup$ – Chee Han Nov 14 '16 at 18:19

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