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Prove that a triangle can be inscribed in the hyperbola $xy=c^2$ whose sides touch the parabola $y^2=4ax$.

I have no idea how to start this problem. In my first attempt I assumed any three points on the given rectangular hyperbola and wrote down the equations of each of the three chords. Then I tried proving that each of the three sides are tangents to the given parabola...then I put the value of $x$ from the equation of the chord in the equation of the parabola and tried to prove that the discriminant of the quadratic is equal to zero...but I am unable to proceed after that.

Need help...thanks!!

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    Are you sure ALL sides of the triangle must touch the parabola? – 2017-02-11
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    I think the parabola touches the lines to which triangle sides belong, rather than the sides themselves. – 2017-02-11

2 Answers 2

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HINT.

From any point $A$ of the hyperbola, in the branch intersecting the parabola but external to it (see diagram), draw tangents $AD$ and $AF$ to the parabola, which intersect the other branch of the hyperbola at $B$ and $C$. Check that line $BC$ is also tangent to the parabola.

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Too long for a comment but I thought I might contribute some formulae

Definitions

  • Pythagoras Thereom $ x^{2} + y^{2} = z^{2} $

hyperbola has co-ordinates $ P = (x, \frac{c^{2}}{y}) $ , $ Q = (\frac{-c^{2}}{x}, y) $ and $ R = (x, \frac{-c^{2}}{y}) $

  • Such points $ P, Q, R $ can be treated as 2D compound numbers that multiply under the composition laws of a field $\mathbb{F} $

  • An equation exists in the form $ y - y_{1} = m(x - x_{1}) $ with $ m $ being the gradient in the form of $ m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} $

  • Simultaneous equations can be solved for intersections points

  • Midpoint of two co-ordinates are $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) $

Proposition

Apply pythagoras thereom to hyperbolic co-ordinates to show that a triangle can be inscribed.

Where such co-ordinates are multiplied out in accord to $ \mathbb{F} $

$ (a,b)(c,d) = (ac + ad, bc + bd) $

$ (x - \frac{c^{2}}{x}, y + \frac{c^{2}}{y})^{2} + (x - x, \frac{c^{2}}{y} - \frac{c^{2}}{y})^{2} = (x - \frac{c^{2}}{x}, y + \frac{c^{2}}{y})^{2} $

Therefore any triangle can be inscribed in a hyperbola, albeit trivial.

The triangle can then be shown to intercept the parabola $ y^{2} = 4ax $ by simultaneous equations.

$ m_{PQ} = \frac{1 - \frac{1}{y}}{\frac{-1}{x} - x} = \frac{yx^{2} - yx}{-yx^{2} -yx} $

using $ y - y_{1} = m(x - x_{1}) $ on $-P $ we get $ y + \frac{1}{y} = \frac{yx^{2} - yx}{-yx^{2} -yx}(x + x) $

simplifies to:

$ 2y^{2} - (4a -1)y + 4ax + \frac{1}{x} = 0 $

$ y - 2y^{2} = \frac{1}{x} $

and has intersection points at $ (\pm \frac{1}{2}, \pm 2) $ and $ (\pm 1, \pm 1) $

Proving a triangle can be embedded by a parametization of $ y^{2} = 4ax $ in $ xy < c^{2} $ resulted in $ x^{2} + y^{2} > xy $ for a right angled triangle which has no solution. I would like to see a proof that such a triangle has to be hyperbolic itself. I may return later this month with appropriate parameter inspection techniques.