0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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*}

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.