#f(x) = sqrt (x^2-1)# and #g(x) = px^2 + 1# touch each other. What's the value of p in this case?

#p epsilon RR#
They touch each other (there are 2 points not four!)

1 Answer
Douglas K.
Dec 11, 2017

#p = 1/(2sqrt(3+2sqrt2)#

Explanation:

#f(x) = sqrt (x^2-1)#
#g(x) = px^2 + 1#

For them to be tangent, they must have the same first derivative:

#f'(x) = x/sqrt(x^2-1)#

#g'(x) = 2px#

Set #g'(x) = f'(x)#:

#2px = x/sqrt(x^2-1)#

#p = 1/(2sqrt(x^2-1)#

Substitute into #g(x)#

#g(x) = x^2/(2sqrt(x^2-1)) + 1#

They must have the same y value, therefore, we may set #f(x) = g(x)#

#sqrt (x^2-1) = x^2/(2sqrt(x^2-1)) + 1#

Multiply both sides by #2sqrt(x^2-1)#:

#2(x^2-1) = x^2+ 2sqrt(x^2-1)#

#2x^2-2 = x^2+ 2sqrt(x^2-1)#

#x^2-2 = 2sqrt(x^2-1)#

Square both sides:

#x^4-4x^2+4 = 4(x^2-1)#

#x^4-4x^2+4 = 4x^2-4#

#x^4-8x^2+8 = 0#

Let #u = x^2#:

#u^2-8u+8#

#u = (8+-sqrt((-8)^2-4(1)(8)))/(2(1))#

#u = 4+-2sqrt2#

Reverse the substitution:

#x^2 = 4+2sqrt2# and #x^2 = 4-2sqrt2#

Substitute into our equation for p:

#p = 1/(2sqrt(3+2sqrt2)# and #p = 1/(2sqrt(3-2sqrt2)#

Here is a graph for the two equations with #p = 1/(2sqrt(3-2sqrt2)#:

www.desmos.com/calculator

They do not touch, therefore, this is an extraneous value.

Here is a graph for the two equations with #p = 1/(2sqrt(3+2sqrt2)#:

www.desmos.com/calculator

They touch, therefore, this is the correct value for p.

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