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What is the shortest/longest distance from $9x^2 + 4y^2 = 36$ to $(5,5)$?

Using Langrange Multipliers, I've set up the standard equation with $$g(x,y) = (x/2)^2 + (y/3)^2 = 1$$ $$f(x,y) = (x-5)^2 + (y-5)^2$$ and $$ \nabla f = -\lambda \nabla g.$$ This gives us $$2(x-5) = - \lambda x / 2$$ $$(y-5) = - \lambda y / 9.$$ Solving for $y$ and $x$, I have $$ y = 9x / (x + 4) $$ $$ x = 4y / (9-y) .$$ But if I plug this value for $y$ into the original ellipse, I get $$ x^4 + 8x^3 + 48x^2 - 32x - 64 = 0.$$ Somehow this doesn't seem quite right as it's now cumbersome to solve for $x$. Where am I going wrong and is there a better approach?

mathjacks
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    Solve for $x$ and $y$ in terms of $\lambda$ first (e.g. $x = \frac{20}{4+\lambda}$) and insert this into the ellipse equation to get a quadratic equation for $\lambda$. – Winther Apr 11 '15 at 15:45
  • With that method, you can't really avoid a quartic (and the answers are a big horrible pile of nested radicals anyway). And @Winther, I had a look at that, and you actually get a quartic for $\lambda$ which looks even worse. – Chappers Apr 11 '15 at 15:46
  • @Chappers I don't think so. Unless I'm missing something, after substituting in, the equation should look like $\frac{a}{(b+\lambda)^2} + \frac{c}{(d+\lambda)^2} = 0$. When you clear denominators you get $a(d+\lambda)^2+c(b+\lambda)^2=0$. – Ian Apr 11 '15 at 15:52
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    @Ian It's an ellipse equation, so isn't there a $1$ on the right-hand side? – Chappers Apr 11 '15 at 16:09

1 Answers1

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We have $$\frac{x^2}4+\frac{y^2}9=1$$

so, any point can be written as $(2\cos t,3\sin t)$

If the distance if $d,$

$$d^2=(2\cos t-5)^2+(3\sin t-5)^2$$

Now use Second derivative test

lab bhattacharjee
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  • I'm torn on this answer. On one hand, it is correct and gives an easier approach than the OP's approach. On the other hand, if the OP's question was intended to be a Lagrange multipliers exercise, then there should be a good way to use Lagrange multipliers to solve it. (Indeed, this approach is essentially the same as solving the constraint for one variable and substituting.) +1 either way. – Ian Apr 11 '15 at 15:40
  • This is a good answer also. Although, I see we want to set $d/dx (d^2) =0$ and solve for $t$. What is the best way to do this by hand? It again seems cumbersome. – mathjacks Apr 11 '15 at 17:52