# Thread: Minimize Area of ellipsis

1. ## Minimize Area of ellipsis

Consider an ellipse of equation $\frac{x^2}{a^2}$+ $\frac{y^2}{b^2}$= $1$ that contains the circle $x^2+y^2=2y$

The question is: Wich values of a and b minimize the area of the ellipse?

2. Originally Posted by orbit
Consider an ellipsis of equation $\frac{x^2}{a^2}$+ $\frac{y^2}{b^2}$= $1$ that contains the circle $x^2+y^2=2y$

The question is: Wich values of a and b minimize the area of the ellpsis?
The area of the ellipse $\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$ is $\pi ab$. Now, what is the circle $x^2+ y^2= 2y$? That is, where is its center and what is its radius? It looks to me like you want your ellipse to just contain that circle which means it must be tangent (at, I think, two points).

3. Originally Posted by orbit
Consider an ellipse of equation $\frac{x^2}{a^2}$+ $\frac{y^2}{b^2}$= $1$ that contains the circle $x^2+y^2=2y$

The question is: Wich values of a and b minimize the area of the ellipse?
Hi orbit.

My guess is that the area of the ellipse is minimized when it’s a circle.

4. A circle passing through (0,2) so $a=b=2.$

5. like HallsofIvy said, "you want your ellipse to just contain that circle which means it must be tangent". In other words, they'll have the same $dy/dx$. This suggests you to take the derivative.

Futhermore, you also need to realize that at that point y value shall be the same.

So, solve $x^2$ in circlue equation in terms of $y$.

Then, plug that value in $x^2$ of the ellispe equation.

Now write the ellipse equation for y in terms of a and b (remember you already eliminated x by substitution).

Next, use implicit differentiation on both equations.

edit: that's an excellent idea TheAbstractionist gave you. Should be lot faster than what i said

6. Originally Posted by TheAbstractionist
A circle passing through (0,2) so $a=b=2.$
I thought the same thing, but I donīt know if it`s alright. Because if we think the ellipse as the circle, the problem is too easy.

Lol

7. Anybody else wants to say something?