the equation of the common tangent to the curves
y^2 = 8x
and
xy = -1
is?

Question

the equation of the common tangent to the curves 
y^2 = 8x 
and
xy = -1
is?

anonymous · Answer

@Astrophysics can you help me out?

anonymous · Answer

so I take a general tangent to one of the curves, then I apply the condition that it is tangent to the other.

anonymous · Answer

as a matter of fact I know that the general tangent of y^2 = 8x is 
y = mx + 2/m

anonymous · Answer

xy = -1
now I've got y' = 1/x^2
what next?

astrophysics · Answer

Ok good so we have $$y=mx+\frac{ 2 }{ m }$$ for $$y^2 = 8x$$ then your second equation $$xy=-1$$ should have the tangent. $$x \left( mx+\frac{ 2 }{ m } \right) = -1$$ find the m

astrophysics · Answer

You can use differentiation if you like but I don't think you need to

anonymous · Answer

you wrote x(mx+2/m) = -1 as the tangent for the second curve but that's not even a line unless I'm missing something
like if m is dependent on x.

astrophysics · Answer

No I mean the tangent also touches the curve xy=-1

anonymous · Answer

oh of course yeah, that makes sense
so how would we go about that?

astrophysics · Answer

$$x \left( mx+\frac{ 2 }{ m } \right) = -1 $$ solving the quadratic will give you the value of your slope here which will give you the tangent for BOTH of these equations once you plug the values in $$y=mx+\frac{ 2 }{ m } $$

anonymous · Answer

will that really work?

parthkohli · Answer

Use discriminants! That quadratic equation will only have one solution. So equate the discriminant to zero.