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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.

 Nov 27, 2018
 #1
avatar+107405 
+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

 

 

cool cool cool

 Nov 27, 2018
 #2
avatar+814 
+1

oops, yeah. Good job, CPhill!

mathtoo  Nov 27, 2018
edited by mathtoo  Nov 27, 2018
 #3
avatar+20231 
0

Welllllll......here is a graph.....

 

ElectricPavlov  Nov 27, 2018
 #4
avatar+107405 
+1

Thanks, EP

 

That looks pretty circular to me.....!!!!!

 

 

cool cool cool

CPhill  Nov 27, 2018
 #5
avatar+814 
+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

 Nov 27, 2018
 #6
avatar+107405 
+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)

 

cool cool cool

CPhill  Nov 27, 2018
edited by CPhill  Nov 27, 2018

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