Please help!

[tpo7]

On this problem:
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. It is possible to draw an ellipse so that it is tangent to all sides of the square; several examples are shown below.
Find necessary and sufficient conditions on a > 0 and b > 0 such that the ellipse
{x^2}/{a^2} + {y^2}/{b^2} = 1 is contained inside the square (and tangent to all of its sides.

My steps so far have been substitutey+x=k into {(x^2)/a^2)}+ {(y^2)/(a^2)} and simpify it down to (b^2+a^2)x^2-2a^2kx+a^2k^2-a^2b^2=0. Can someone show me what to do from here? I really, really appreciate the help!

 

[HallsofIvy]

Note that the sides of the square have slopes 1 (the sides from (0,-k) to (k, 0) and from (-k, 0) to (0, k)) and -1 (the sides from (0, -k) to (-k, 0) and from (0,k) to (k,0)).

The ellipse $\displaystyle \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$ has derivative given by $\displaystyle \frac{2x}{a^2}+ \frac{2yy'}{b^2}= 0$ so $\displaystyle y'= -\frac{b^2x}{a^2y}$. Where is that 1 or -1?

 

[tpo7]

Thank you! I actually solved this problem on my own, but thanks for your input!

 

[jonah2.0]

Beer soaked suggestion follows.

Thank you! I actually solved this problem on my own, but thanks for your input!

Can you post your work?
Others can learn from it.

 

[tpo7]

Here it is:
Alright, let's start by looking at the ellipse formula {x^2}/{a^2} + {y^2}/{b^2} = 1. It is given that a line tangent to this is x + y = k. The best possible way to do this is substitution.
We can set this as y = k - x and substitute it into {x^2}/{a^2} + {y^2}/{b^2} = 1.

{x^2}/{a^2}+\{(-x+k)^2}{b^2}=1
Now, we multiply it by a^2b^2.
{x^2}/{a^2}a^2b^2+{(-x+k)^2}/{b^2}a^2b^2=1a^2b^2
Then we simplify:
b^2x^2+a^2(-x+k)^2=a^2b^2
Now to expand:
b^2x^2+a^2(-x+k)^2
b^2x^2+a^2x^2-2a^2kx + a^2k^2
b^2x^2+a^2x^2-2a^2kx+a^2k^2=a^2b^2
Now, we subtract a^2b^2 from both sides:
b^2x^2+a^2x^2-2a^2kx+a^2k^2-a^2b^2=a^2b^2-a^2b^2
Now to simplify further:
(b^2+a^2)x^2-(2a^2*k)x+(a^2k^2-a^2b^2)=0


This equation ((b^2+a^2)x^2-2a^2kx+a^2k^2-a^2b^2=0) is a quadratic equation.
Let us find the discriminant of this.
Now, a discriminant is b^2-4ac. This looks much more intimidating than it actually is to find the discriminant.
Our “a” would be b^2+a^2, our “b” would be -2a^2*k, and our “c” would be a^2k^2 - a^2b^2.

Let's find the discriminant:
(-2a^2*k)^2-4(b^2+a^2)(a^2k^2-a^2b^2)
(-2a^2*k)^2 = 2^2a^4k^2
2^2a^4k^2-4(a^2+b^2)(a^2k^2-a^2b^2)
=4a^4k^2-4(a^2+b^2)(a^2k^2-a^2b^2)
Now we have to expand -4(a^2+b^2)(a^2k^2-a^2b^2):
-4(b^2 + a^2)(a^2k^2-a^2b^2) = -4a^2b^2k^2 + 4a^2b^4 - 4a^4k^2 + 4a^4b^2

Now we have 4a^4k^2-4a^2b^2k^2+4a^2b^4-4a^4k^2+4a^4b^2
Time for us to simplify this out!
After simplifying, we get 4a^4b^2+4a^2b^4-4a^2b^2k^2.

We can use the Exponent Rule to get this equation: 4a^2a^2b^2+4a^2b^2b^2-4a^2b^2k^2
Now, we factor out the common term of 4b^2a^2, to get the equation of
4b^2a^2(a^2+b^2-k^2).
Now, the discriminant must be 0 in order for there to be only one solution
where the ellipse just touches the square. So, our next step would be to set the discriminant equal to 0.
4b^2a^2(a^2+b^2-k^2) = 0


(Someone can probably go on on their own from here)

 

[jonah2.0]

Beer soaked suggestion follows.

It's been revealed recently (by Dr. Peterson) here that enclosing a mathematical expression with a symbol whose name I can't recall at the moment (that backslash version of ') will cause them to be shown in latex format. So,

tpo7 said:
Here it is:
Alright, let's start by looking at the ellipse formula ${\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$. It is given that a line tangent to this is ${\displaystyle x+y=k}$. The best possible way to do this is substitution.
We can set this as ${\displaystyle y=k-x}$ and substitute it into ${\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$.

${\displaystyle \frac{x^2}{a^2}+\left\{{(-x+k)}^2\right\}\left\{b^2\right\}=1}$
Now, we multiply it by ${\displaystyle a^2{b}^2}$.
${\displaystyle \frac{x^2}{a^2}{a}^2{b}^2+\frac{{(-x+k)}^2}{b^2}{a}^2{b}^2=1a^2{b}^2}$
Then we simplify:
${\displaystyle b^2{x}^2+a^2{(-x+k)}^2=a^2{b}^2}$
Now to expand:
${\displaystyle b^2{x}^2+a^2{(-x+k)}^2}$
${\displaystyle b^2{x}^2+a^2{x}^2-2a^{2}kx+a^2{k}^2}$
${\displaystyle b^2{x}^2+a^2{x}^2-2a^{2}kx+a^2{k}^2=a^2{b}^2}$
Now, we subtract ${\displaystyle a^2{b}^2}$ from both sides:
${\displaystyle b^2{x}^2+a^2{x}^2-2a^{2}kx+a^2{k}^2-a^2{b}^2=a^2{b}^2-a^2{b}^2}$
Now to simplify further:
${\displaystyle (b^2+a^2)x^2-(2a^2\cdot k)x+(a^2{k}^2-a^2{b}^2)=0}$


This equation ${\displaystyle ((b^2+a^2)x^2-2a^{2}kx+a^2{k}^2-a^2{b}^2=0)}$ is a quadratic equation.
Let us find the discriminant of this.
Now, a discriminant is ${\displaystyle b^2-4ac}$. This looks much more intimidating than it actually is to find the discriminant.
Our “${\displaystyle a}$” would be ${\displaystyle b^2+a^2}$, our “${\displaystyle b}$” would be ${\displaystyle -2a^2\cdot k}$, and our “${\displaystyle c}$” would be ${\displaystyle a^2{k}^2-a^2{b}^2}$.

Let's find the discriminant:
${\displaystyle {(-2a^2\cdot k)}^2-4(b^2+a^2)(a^2{k}^2-a^2{b}^2)}$
${\displaystyle {(-2a^2\cdot k)}^2=2^2{a}^4{k}^2}$
${\displaystyle 2^2{a}^4{k}^2-4(a^2+b^2)(a^2{k}^2-a^2{b}^2)=4a^4{k}^2-4(a^2+b^2)(a^2{k}^2-a^2{b}^2)}$
Now we have to expand ${\displaystyle -4(a^2+b^2)(a^2{k}^2-a^2{b}^2)}$:
${\displaystyle -4(b^2+a^2)(a^2{k}^2-a^2{b}^2)=-4a^2{b}^2{k}^2+4a^2{b}^4-4a^4{k}^2+4a^4{b}^2}$

Now we have ${\displaystyle 4a^4{k}^2-4a^2{b}^2{k}^2+4a^2{b}^4-4a^4{k}^2+4a^4{b}^2}$
Time for us to simplify this out!
After simplifying, we get ${\displaystyle 4a^4{b}^2+4a^2{b}^4-4a^2{b}^2{k}^2}$.

We can use the Exponent Rule to get this equation: ${\displaystyle 4a^2{a}^2{b}^2+4a^2{b}^2{b}^2-4a^2{b}^2{k}^2}$
Now, we factor out the common term of ${\displaystyle 4b^2{a}^2}$, to get the equation of
${\displaystyle 4b^2{a}^2(a^2+b^2-k^2)}$.
Now, the discriminant must be 0 in order for there to be only one solution
where the ellipse just touches the square. So, our next step would be to set the discriminant equal to 0.
${\displaystyle 4b^2{a}^2(a^2+b^2-k^2)=0}$


(Someone can probably go on on their own from here)

Note: I see a minor typo.