Proof that this system is tangent to a parabola

[Uranium235]

Homework Statement

Well I would like to prove that any equation that follows the pattern y=rx-r^(-1) is tangent to some sideways parabola (I know this to be true). Problem is that I need help in finding the parabola in question and actually proving my conjecture. I do know, after graphing, that the parabola has a vertex a (0,0).

Homework Equations

y=rx-r^(-1) any linear system of this form should be tangent to a sideways parabola
y=(ax)^1/2 the equation of a sideways parabola

The Attempt at a Solution

After graphing, I realized that any equation of the form y=rx-r^(-1) where r is positive seems to be always tangent the curve y=-(-4x)^(1/2) while any equation of the form y=rx-r^(-1) where r is negative seems to be always tangent to y=(-4x)^(1/2). I got those values from trial and error only and there is no proof to support it.

Any help in clearing this up will be appreciated since my assignment is due on friday. Thank you!

 

[matiasmorant]

you are saying that y=(ax)^1/2, So y^2=ax. Let`s find all it's tangent lines, (that is let's find the tangen in the point X Y)
Differentiating y^2=ax with respect to x gives 2 y = a dx/dy, so dy/dx=a/2y. The tangent line must have this slope. and also the point {X,Y} must be included, so "b" must stisfy the equation
Y=dy/dx X + b so b=Y-Xdy/dx thus, the equation of the tangent line at the point {X,Y} is

y=a/2Y x +Y-Xa/2Y

So, what you are saying is true as long as r=a/2Y and r^-1=Xa/2Y-Y
the first equation is the same as r^-1 =2Y/a
So, what you are saying is true as long as Xa/2Y-Y=2Y/a
Simplifying a little yields X=(4/a^2 + 2/a) Y^2
and, since Y^2=a X, then, the last equation is
X=(4/a^2 + 2/a) aX
which holds for any value of X, so
1=4/a + 2
a=-4 which is what you were looking for.

Your equations is, then y^2=-4 x. This is, of course, a parabola in the half plane of negative x, since y would have a complex value if x were positive.