Elliptic Triangles

As promised, today I have a question involving ellipses. In my opinion, it's a great question.

C is a point on the ellipse x²/16 + y²/9 = 1. A and B are the foci of the ellipse. Prove that angle ∠ACB cannot be 90°.

To do that, we first need to find the foci of the ellipse. As you remember, in an ellipse, a² = b² + c². From the equation of the ellipse, a² = 16 and b² = 9. This means:
c² = a² - b² = 16 - 9 = 7
c = √7

Now that we know c, we know one side of triangle ABC. The side is AB and it equals 2√7. Now we only have to find the other sides and show that ∠ACB cannot equal 90°.

Let's call one side, AC, x. By the definition of an ellipse, the sum of the distances from a point on the ellipse to the foci is a constant, 2a. In this case, 2a = 2√16 = 8. This means the third side, BC, is 8 - x.

We have three sides of the triangle and we want to find an angle. To do that, we can use the law of cosines, which involves all three sides and an angle. Since we want ∠ACB, we will use the side opposing it, AB.

The law of cosines states that:
a² = b² + c² - 2bc * cos∠A
(a, b, c, and ∠A are not related to the ellipse)

Let's plug our known values in there:
(2√7)² = x² + (8 - x)² - 2x(8 - x)cos∠ACB

We want to know what happens when ∠ACB = 90°, which means cos∠ACB = 0. Let's do all the squaring and eliminate the last part, because it's zero:
28 = x² + 64 - 16x + x²
2x² - 16x + 36 = 0
x² - 8x + 18 = 0

When trying to solve this through the quadratic formula, the expression inside the square root equals:
8² - 4*18 = 64 - 72 = -8

There is no real number for x which makes the law of cosines correct for 90°, so angle ∠ACB cannot equal 90° in this ellipse.

There are even cooler things on Super Math Tips, go check them out.

Hope you liked it,
Nadav

nadavs