<PRE><B>Question 63902</B>
You need to complete the square to get this equation into standard form and then see what it matches up to.  &lt;br&gt;

x^2 - 4y^2 + 2x - 8y - 10 = 0&lt;br&gt;

x^2 + 2x - 4(y^2 - 2y) = 10&lt;br&gt;

x^2 + 2x + 1 - 4(y^2 - 2y + 1) = 10+4+1&lt;br&gt;

(x+1)^2 - 4(y-1)^2 = 15&lt;br&gt;

((x+1)^2)/15 - (4(y-1)^2)/15 = 1&lt;br&gt;

I can&#39;t remember if this is the standard form for an ellipse or a hyperbola but it is one of the two.</PRE>