Ellipse with rectangular frame

**courteous** · January 24th 2009, 04:49 AM

»Write down the equation of an ellipse, that goes through $P(1,-2)$ and whose circumscribed* rectangle's area is minimized.«
*

(Pretending not knowing that "rectangle" is in fact square.)

Set-up:
Ellipse through $(1,-2)$ : $\frac{1}{a^2}+\frac{4}{b^2}=1$ , rectangle's area: $2a*2b=min.$
Now, what to differentiate? From the first equation I've also got $a^2b^2=b^2+4a^2$ .

**skeeter** · January 24th 2009, 06:52 AM

$a^2b^2 = b^2 + 4a^2$

$a^2b^2 - 4a^2 = b^2$

$a^2(b^2 -4) = b^2$

$a^2 = \frac{b^2}{b^2 - 4}$

$a = \frac{b}{\sqrt{b^2 - 4}}$

$A = 4ab$

$A = \frac{4b^2}{\sqrt{b^2 - 4}}$

find $\frac{dA}{db}$ and determine the value of b that minimizes the area.

**Danny** · January 24th 2009, 08:13 AM

Originally Posted by courteous

»Write down the equation of an ellipse, that goes through $P(1,-2)$ and whose circumscribed* rectangle's area is minimized.«
*

(Pretending not knowing that "rectangle" is in fact square.)

Set-up:
Ellipse through $(1,-2)$ : $\frac{1}{a^2}+\frac{4}{b^2}=1$ , rectangle's area: $2a*2b=min.$
Now, what to differentiate? From the first equation I've also got $a^2b^2=b^2+4a^2$ .

You could use Lagrange multipliers. Minimizing $A = 4ab$ subject to $\frac{1}{a^2} + \frac{4}{b^2} = 1$ is to use

$F = 4ab + \lambda \left( \frac{1}{a^2} + \frac{4}{b^2} - 1\right)$

then calculate $F_a,\;\;\;F_b,\;\;\;F_{\lambda}$ , set them to zero and solve for $a, b, \; \text{and}\; \lambda$ . It avoids the square roots. (You sure about the square?)

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