Question

What are the foci of the ellipse given by 3x^2+4y^2=48

Answer 1

First you need to divide both sides by 48 to get the equation in standard form \[\frac{3x^2}{48} + \frac{4y^2}{48} = 1\] Now you need to simplify \[\frac{x^2}{16} + \frac{y^2}{12} = 1\]. Now you need to use the formula

\[c^2 = a^2 - b^2\] where c is the distance from the center to one of the foci, a^2 = 16 and b^2 = 12

so c^2 = 4 so c = + - 2 so the foci are at (2,0) and (-2,0)