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

May 13, 2016

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

#### Explanation:

Function $f$

$f \left(x\right) = 6 - 10 {x}^{2}$ so $f ' \left(x\right) = - 20 x$

At a point $\left(a , f \left(a\right)\right)$, the equation of the tangent line is

$y = f \left(a\right) + f ' \left(a\right) \left(x - a\right)$

$= \left(6 - 10 {a}^{2}\right) + \left(- 20 a\right) \left(x - a\right)$ which can be simplified to

$y = - 20 a x + 10 {a}^{2} + 6$

Function $g$

$g \left(x\right) = 4 + 4 x - {x}^{2}$ so $g ' \left(x\right) = 4 - 2 x$

At a point $\left(b , g \left(b\right)\right)$, the equation of the tangent line is

$y = g \left(b\right) + g ' \left(b\right) \left(x - b\right)$

$= \left(4 + 4 b - {b}^{2}\right) + \left(4 - 2 b\right) \left(x - b\right)$ which can be simplified to

$y = \left(4 - 2 b\right) x + {b}^{2} + 4$

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

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

$10 {a}^{2} + 6 = {b}^{2} + 4 = {\left(10 a + 2\right)}^{2} + 4$

$90 {a}^{2} + 40 a + 2 = 0$

The solutions are $a = \frac{- 10 \pm \sqrt{55}}{45}$ and $b = 10 a + 2$

Using the approximations $a \approx - 0.0374178$ and $a \approx - 0.387027$

we get corresponding values of $b \approx 1.87027$ and $b \approx - 1.87027$

Using these in $y = - 20 a x + 10 {a}^{2} + 6$ or $y = \left(4 - 2 b\right) x + {b}^{2} + 4$ gets us the lines

$y = 1.148 x + 6.033$ and

$y = 7.741 x + 7.498$

May 13, 2016

There are two solutions as detailed below

#### Explanation:

The tangent lines to $f$ and $g$ are given by
${t}_{f} \to \left(y - f \left(a\right)\right) = {f}_{x} \left(a\right) \left(x - a\right)$
${t}_{g} \to \left(y - g \left(b\right)\right) = {g}_{x} \left(b\right) \left(x - b\right)$ where ${f}_{x} , {g}_{x}$ indicates derivatives with respect to $x$.
${t}_{f} \to y - 6 + 10 {a}^{2} = - 20 a \left(x - a\right)$
${t}_{g} \to y - 8 + {\left(b - 2\right)}^{2} = - 2 \left(b - 2\right) \left(x - b\right)$
Equating the $y$ values
$2 \left(3 + 5 {a}^{2} - 10 a x\right) = 4 + {b}^{2} + \left(4 - 2 b\right) x$
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:
$\left\{a \to \frac{1}{45} \left(- 10 - \sqrt{55}\right) , b \to \frac{2}{9} \left(- 1 - \sqrt{55}\right)\right\}$ and
$\left\{a \to \frac{1}{45} \left(- 10 + \sqrt{55}\right) , b \to \frac{2}{9} \left(- 1 + \sqrt{55}\right)\right\}$