Which type of conic section is described by the equation\(\frac{1}{x^2} + \frac{1}{y^2} = \frac{x+4}{x^2y^2}.\)Enter "C" for circle, "P" for parabola, "E" for ellipse, "H" for hyperbola, and "N" for none of the above.

mathtoo Nov 27, 2018

#1**+2 **

Multiply both sides by x^2y^2 and we get

y^2 + x^2 = x + 4 rearrange as

x^2 - 1x + y^2 = 4 complete the square on x

x^2 - 1x + 1/4 + y^2 = 4 + 1/4 factor the first three terms

(x - 1/2)^2 + (y - 0)^2 = 17/4

This is a circle centered at (1/2, 0 ) and a radius of sqrt (17) / 2

CPhill Nov 27, 2018

#5**+1 **

Wait, this circle intersects the two axes at several points, whereas no point with x=0 or y=0 lies on the graph of the original equation

mathtoo Nov 27, 2018

#6**+1 **

They are trying to trick you, mathtoo....it's a "disguised" conic

For instance...look at this graph : https://www.desmos.com/calculator/kdzjzqgkr4

It appears that for the function y / x^2 = 3 the point (0,0) couldn't be on the graph

But....rearranging the equation as y = 3x^2 shows that it is a parabola with a vertex at (0,0)

CPhill
Nov 27, 2018