Question

Do `f(x) = 6 – 10x^2` and `g(x) = 8 – (x – 2)^2` share any tangent lines?

Answer 1

Yes, they do. The lines are (approximately) `y=7.741x+7.498` and `y=1.148x+6.033`

Explanation:

Function `f`

`f(x) = 6-10x^2` so `f'(x) = -20x`

At a point `(a,f(a))`, the equation of the tangent line is

`y=f(a)+f'(a)(x-a)`

`= (6-10a^2)+(-20a)(x-a)` which can be simplified to

`y= - 20ax+10a^2+6`

Function `g`

`g(x) = 4+4x-x^2` so `g'(x) = 4-2x`

At a point `(b,g(b))`, the equation of the tangent line is

`y=g(b)+g'(b)(x-b)`

`= (4+4b-b^2)+(4-2b)(x-b)` which can be simplified to

`y = (4-2b)x +b^2+4`

The lines coincide when their slopes are the same and their `y` intercepts are the same.

`-20a = 4-2b` which implies `b=10a+2`

`10a^2+6=b^2+4 = (10a+2)^2+4`

This leads to

`90a^2+40a+2=0`

The solutions are `a=(-10+-sqrt(55))/45` and `b=10a+2`

Using the approximations `a ~~ -0.0374178` and `a ~~ -0.387027`

we get corresponding values of `b ~~ 1.87027` and `b ~~ -1.87027`

Using these in `y= - 20ax+10a^2+6` or `y = (4-2b)x +b^2+4` gets us the lines

`y=1.148x+6.033` and

`y=7.741x+7.498`

Answer 2

There are two solutions as detailed below

Explanation:

The tangent lines to `f` and `g` are given by
`t_f->(y-f(a))=f_x(a)(x-a)`
`t_g->(y-g(b))=g_x(b)(x-b)` where `f_x,g_x` indicates derivatives with respect to `x`.
the equations read
`t_f->y-6 + 10 a^2 = -20 a (x-a)`
`t_g->y-8 + (b-2)^2 = -2 (b-2) (x-b)`
Equating the `y` values
`2 (3 + 5 a^2 - 10 a x)=4 + b^2 + (4 - 2 b) x`
and imposing the equality for all `x` we obtain the conditions
`2 + 10 a^2 - b^2=0,20 a + 4-2 b=0`
Solving for `a,b` we obtain two solutions:
`{a -> 1/45 (-10 - sqrt[55]), b -> 2/9 (-1 - sqrt[55])}` and
`{a -> 1/45 (-10 + sqrt[55]), b -> 2/9 (-1 + sqrt[55])}`