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# QUESTION ABOUT CONICS PLS HELP

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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

### 9+0 Answers

#1
+1

What's an elliipse?

Apr 29, 2020
#2
+111082
+1
Melody  Apr 29, 2020
#3
+25569
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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
$$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$
is contained inside the square (and tangent to all its sides).

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

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

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

Apr 29, 2020
#4
+111082
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Thanks Heureka,

For this answer but also for all your wonderful answers.

Apr 29, 2020
#5
+25569
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Thank you very much, Melody !

heureka  Apr 30, 2020
#6
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Heureka, what do you do to get to $$-\frac{b^2}{a^2}\frac{x_t}{y_t}=1$$

May 27, 2020
#7
+25569
+3

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

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

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