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Let k be a positive real number. The square with vertices (k, 0), (0,k), (-k,0), and (0,-k) is plotted in the coordinate plane. 

 

Find conditions on a > 0 and b > 0 such that the ellipse 

 

\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

 

is contained inside the square (and tangent to all its sides). 

 Apr 29, 2020
 #1
avatar
+1

What's an elliipse?

 Apr 29, 2020
 #2
avatar+111082 
+1
Melody  Apr 29, 2020
 #3
avatar+25569 
+4

Let k be a positive real number.
The square with vertices \((k, 0)\), \((0,k)\), \((-k,0)\), and \((0,-k)\) is plotted in the coordinate plane.

 

Find conditions on \(a > 0\) and \(b > 0\) such that the ellipse
\(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\)
is contained inside the square (and tangent to all its sides).

 

\(\text{Let $x_t,y_t$ are points on the tangent line $\mathbf{y = x+k}$ from $(-k,0)$ to $(0,k)$, so $ \mathbf{y_t = x_t + k}$}\)

 

\(\begin{array}{|lrcll|} \hline \text{ellipse} : & \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} &=& 1 \\ \hline \text{tangent of the ellipse} : & \dfrac{x_tx}{a^2} + \dfrac{y_ty}{b^2} &=& 1 \\ & \dfrac{y_ty}{b^2} &=& 1-\dfrac{x_tx}{a^2} \\ & \ldots \\ & \mathbf{y} &=& \mathbf{\dfrac{b^2}{y_t}-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}x}\\ &&& \boxed{\text{compare with tangent line }\\ y=k+x} \\ 1. & k &=& \dfrac{b^2}{y_t} \ \text{or}\ \mathbf{b^2} = \mathbf{ky_t} \\\\ 2. & 1 &=& -\dfrac{b^2}{a^2}\dfrac{x_t}{y_t} \quad | \quad \mathbf{b^2} = \mathbf{ky_t} \\\\ & 1 &=& -\dfrac{ky_t}{a^2}\dfrac{x_t}{y_t} \\\\ & 1 &=& -\dfrac{kx_t}{a^2} \\\\ & a^2 &=& -kx_t \quad | \quad x_t = y_t -k \\ & a^2 &=& -k(y_t -k) \\ & a^2 &=& -ky_t +k^2 \\ & \mathbf{a^2} &=& \mathbf{k^2-ky_t} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1): & \mathbf{b^2} &=& \mathbf{ky_t} \\\\ (2): & \mathbf{a^2} &=& \mathbf{k^2-ky_t} \quad | \quad \mathbf{b^2=ky_t} \\ & a^2 &=& k^2-b^2 \\ & \mathbf{a^2+b^2} &=& \mathbf{k^2} \\ & \mathbf{\sqrt{a^2+b^2}} &=& \mathbf{k} \\ \hline \end{array} \)

 

 

laugh

 Apr 29, 2020
 #4
avatar+111082 
+1

Thanks Heureka,

For this answer but also for all your wonderful answers.   laugh

 Apr 29, 2020
 #5
avatar+25569 
+2

Thank you very much, Melody !

 

laugh

heureka  Apr 30, 2020
 #6
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0

Heureka, what do you do to get to \(-\frac{b^2}{a^2}\frac{x_t}{y_t}=1\)

 May 27, 2020
 #7
avatar+25569 
+3

Heureka, what do you do to get to \(-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}=1\)

 

\(\begin{array}{|l|} \hline \mathbf{y} = \mathbf{\dfrac{b^2}{y_t}-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}x}\quad \boxed{\text{compare with tangent line }\\ y=k+1x} \\ \hline \end{array} \\ \begin{array}{rclcrl} y&=&\dfrac{b^2}{y_t}-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}x &\color{darkgrey}\text{compare}& y&=&k+1*x \\\\ y&=& {\color{red}\dfrac{b^2}{y_t}}{\color{green}-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}}x &\color{darkgrey}\text{compare}& y&=&{\color{red}k}{\color{green}+1}*x \\ &&&\begin{array}{|c|} \hline {\color{red}\dfrac{b^2}{y_t}} = {\color{red}k} \\\\ {\color{green}-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}} = {\color{green}+1}\\ \hline \end{array} \end{array}\\\)

 

 

laugh

heureka  May 27, 2020
edited by heureka  May 28, 2020
 #8
avatar
+1

Heureka, could you please link the graph in this question because I'd like to see the bottom part of the graph as well.

Guest May 28, 2020
 #9
avatar+111082 
0

Heureka would most likely have to create that graph from scratch.

 

He has given you most of the equations. You should be able to create it for yourself.

 

https://www.desmos.com/calculator

Melody  May 28, 2020

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