Question

`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!)

Answer 1

`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.