Question

would someone please show me how to find the foci of the ellipse: 1x^2+81y^2=25

Answer 1

First, write the equation in the form\[\frac{ x^2 }{ a^2 }+\frac{ y^2 }{ b^2 }=1\]In your ellipse this would mean: divide by 25:\[\frac{ x^2 }{ 5^2 }+\frac{ 9^2 y^2 }{ 5^2 }=1\]If we rewrite this as:\[\frac{ x^2 }{ 5^2 }+\frac{ y^2 }{ \frac{ 5^2 }{ 9^2 } }=1\]or:\[\frac{ x^2 }{ 5^2 }+\frac{ y^2 }{ \left( \frac{ 5 }{ 9 } \right)^2 }=1\]This gives us the numbers\[a=5\]\[b=\frac{ 5 }{ 9 }\]What kind of ellipse is it?
Well, if you set y = 0, x can be 5 or -5, so (-5, 0) and (5, 0) are on the ellipse.
For x = 0, y can be 5/9 or -5/9, so (0, -5/9) and (0, 5/9) also lie on it.

From these points, we see that it is a "horizontal" ellipse.
The foci have coordinates (-c, 0) and (c, 0).
Now \[c^2=a^2-b^2\]So\[c^2=25-\frac{ 25 }{ 81 }=\frac{ 2000 }{ 81 }\]Taking the root gives\[c=\pm \frac{ 20\sqrt{5} }{ 9 }\]and the coordinates of the foci are\[(\pm \frac{ 20\sqrt{5} }{ 9 },0)\]