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# Ellipses

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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 of its sides).

A hint came:

Suppose that the line $$x+y=k$$ is tangent to the ellipse

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$.

Algebraically, what can we say about the solutions? In particular, the number of solutions.

Sorry for the big picture...

Mar 6, 2020

#1
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I like the big picture, and the title.

I'm not good on ellipses though.......

Mar 7, 2020
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What should I do?

#3
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I've given it lots of thought.

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$           and you know that a tangent to this is     y+x=k

Substitute  y=-x+k   into the ellipse formula and simplify it.

It will simplify down to     $$(a^2+b^2)x^2+(-2a^2k)x+(a^2k^2-a^2b^2)=0\\$$

Now this is a quadratic but since there is only one point of intersection (since one is a tangent to the other) there can be only 1 solution.

So that means the discriminant must equal =0

so find the discriminant and set it equal to 0.

Solve it

You will end up with     $$a^2+b^2=k^2$$

Since a, b and k are all positive

The restriction on a and b is          $$\bf{k=\sqrt{a^2+b^2}}$$

Here is the Desmos graph:      https://www.desmos.com/calculator/p2ywsrcrfu

Mar 8, 2020
#4
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Melody, I actually finished the problem and just forgot to tell you, and I got the same answer. Thanks for your imput though!!