Question

What are the foci of the ellipse given by the equation 100x^2 + 64y^2 = 6 400?

Answer 1

The equation of our ellipse is:

$100 x^2 + 64 y^2 = 6400$ (1)

First, let's reduce the equation of the ellipse to the standard form:

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

To do that, we should divide both terms of equation (1) by 6400, and we get:

$\frac{x^2}{64}+ \frac{y^2}{100}=1$

This is a vertical ellipse (because $b^2 \ \textgreater \ a^2$) centered in the origin, and so the distance of its foci from the origin (on the y axis) is given by

$c= \sqrt{b^2-a^2}= \sqrt{(100)-(64)}=6$

Therefore, the position of the two foci is (0,6) and (0,-6)