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Is it a conic? (Posted on 2017-04-19) Difficulty: 3 of 5
You've probably seen this shape before. It is several straight lines whose outline seems to be curved. Connect the following pairs of points with segments:

(1,0) and (0,8)
(2,0) and (0,7)
(3,0) and (0,6)
...
(8,0) and (0,1)

The 8 segments seem to form the outline a curve, but what curve? Is it a circle or maybe one branch of a hyperbola? Some other conic? Does it's equation have some other nice form?

No Solution Yet Submitted by Jer    
Rating: 4.0000 (3 votes)

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Solution Another approach | Comment 6 of 7 |
All eight lines Jer specifies have the sum of their x- and y-intercepts equal to 9.  These can be expressed using a parameter n=1 to 8: (9-n)*x + n*y = n*(9-n).  Letting n vary over all reals creates a set of lines all outside of the curve.

Treat the equation as quadratic in n, which rearranged yields: n^2 + (-9-x+y)*n + (9x) = 0
Which implies n = ((9+x-y) +/- sqrt[(9+x-y)^2-36x])/2

If the discriminant is positive then two values of n represent the two different linear equations containing point (x,y).
If the discriminant is negative then there is no line containing the point (x,y).
Finally, if the discriminant is zero then that point is on the curve and the line it is on is tangent to the curve at that point.

Thus we can define the curve as (9+x-y)^2-36x = 0 and locus of all lines as (9+x-y)^2-36x >= 0.

So what conic is it? First simplify to standard form: x^2 - 2xy + y^2 - 18x - 18y + 81 = 0
Then use the discriminant D = b^2-4*a*c (D>0 implies hyperbola, D=0 implies parabola, D<0 implies ellipse): D = (-2)^2-4*1*1 = 0.
Because D=0, we have a parabola.

  Posted by Brian Smith on 2017-04-20 23:41:14
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