Let
be a positive real number. The square with vertices
and
is plotted in the coordinate plane. It is possible to draw an ellipse so that it is tangent to all sides of the square.
Find necessary and sufficient conditions on
and
such that the ellipse
![\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]](https://latex.artofproblemsolving.com/7/2/f/72febbaca4b25ba0582b718cf7ba5c8b8c6ff29d.png "\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]")
is tangent to all sides of the square.
Make sure to prove that your conditions are both necessary (
) and sufficient (
).
Suppose that the line
is tangent to the ellipse
![\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.\]](https://latex.artofproblemsolving.com/3/d/6/3d6b66349a51e202aca121c17e5d304929dbf9f5.png "\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.\]")
My answer so far:
We can begin by solving the the system x^2/a^2 + y^2/b^2 = 1, x+y = k. Then we can subsitute y = -x + k and simplyifing the equation. This results in us having (a^2+b^2)x^2+(-2a^2k)x+(a^2k^2-a^2b^2) = 0.
The discriminant of this equation must be 0 for there to be only one solution where the ellipse touches each side of the square at only one point. We know that the discriminant is b^2-4ac. Substituting gives us (-2a^2k)^2-4(b^2+a^2)(a^2k^2-a^2b^2). This simplifies to 4a^2b^2(a^2+b^2-k^2). Solving this equation for k gives us k^2 = a^2+b^2.
Since a, b, and k are all positive, k = sqrt(a^2+b^2) is the necessary and sufficient condition such that the ellipse x^2/a^2 + y^2/b^2 = 1, is tangent to all sides of the square.
I need help proving my condition is sufficient. Help is greatly appreciated.
Find necessary and sufficient conditions on
is tangent to all sides of the square.
Make sure to prove that your conditions are both necessary (
Suppose that the line
My answer so far:
We can begin by solving the the system x^2/a^2 + y^2/b^2 = 1, x+y = k. Then we can subsitute y = -x + k and simplyifing the equation. This results in us having (a^2+b^2)x^2+(-2a^2k)x+(a^2k^2-a^2b^2) = 0.
The discriminant of this equation must be 0 for there to be only one solution where the ellipse touches each side of the square at only one point. We know that the discriminant is b^2-4ac. Substituting gives us (-2a^2k)^2-4(b^2+a^2)(a^2k^2-a^2b^2). This simplifies to 4a^2b^2(a^2+b^2-k^2). Solving this equation for k gives us k^2 = a^2+b^2.
Since a, b, and k are all positive, k = sqrt(a^2+b^2) is the necessary and sufficient condition such that the ellipse x^2/a^2 + y^2/b^2 = 1, is tangent to all sides of the square.
I need help proving my condition is sufficient. Help is greatly appreciated.
