# Common tangent lines of two quadratic functions

Find all such lines that are tangent to the following curves:

$$y=x^2$$ and $$y=-x^2+2x-2$$

I have been pounding my head against the wall on this. I used the derivatives and assumed that their derivatives must be equal at those tangent point but could not figure out the equations. An explanation will be appreciated.

Here is a hint for a method which avoids calculus:

The line $y=ax+b$ is a tangent to a quadratic such as $y=x^2$ if and only if the quadratic equation you get by solving these equations simultaneously has a double root. This will give you an equation which must be satisfied by the unknowns $a$ and $b$.

You can do the same for the line $y=ax+b$ and your other quadratic, then solve the two simultaneous equations to find $a$ and $b$.

If $y=x^2$ we have that $\frac {dy}{dx} = 2x$

This gives the gradient of the tangent at the point $(x, y)=(a, a^2)$

If the tangent line is $y=mx + c$ we therefore have $a^2=(2a)\cdot a+c$ whence $c=-a^2$ and the general tangent line to $y=x^2$ is $$y=2ax-a^2$$

If $y=-x^2+2x-2$ we have $\frac {dy}{dx} = -2x+2$, so that the tangent line $y=mx +c$ at $(x, y)=(b, -b^2+2b-2)$ is $-b^2+2b-2=(-2b+2)b+c$ whence $c=b^2-2$ and the general tangent to the parabola is $$y=(-2b+2)x+b^2-2$$

If the two tangents are to be the same line we equate coefficients to give:$$a=1-b$$ and $$a^2=2-b^2$$

Tangent line of first equation through some point $(x_1,f_1(x_1))$ is $$y = f_1(x_1) + f'_1(x_1)(x-x_1) = x_1^2 + 2x_1(x-x_1) = 2x_1x-x_1^2$$ Tangent line of second equation through some point $(x_2, f_2(x_2))$ is $$y = f_2(x_2) + f'_2(x_2)(x-x_2) = -x_2^2+2x_2-2 + (-2x_2+2)(x-x_2) = 2(1-x_2)x+x_2^2-2$$ In order these two lines to be the same one must require $$2x_1 = 2(1-x_2) \\ -x_1^2 = x_2^2-2$$ It is quite simple to solve so I leave it to you. Solution is $$k = 1 \pm \sqrt 3 \\ b = -1 \pm \frac {\sqrt 3}2$$ for the line $y = kx + b$.

• Thanks for this answer. There is something that I do not understand: If you equal both 1st and 2nd equations, you reach to the following equation: $$2x_1x-x_1^2 = 2(1-x_2)x+x_2^2-2$$ which is one equation and two unknowns ($x$ and $x_2$). How do you solve it? I would appreciate very much if you could answer me please. Thanks a lot – DavidC. Oct 18 '17 at 13:12
• @DavidC. hey, sorry for the late response. I think you misunderstood the statement, you don't have to solve the equation you wrote. Instead you need to find such values of $x_1$ and $x_2$ that make both sides the same. You do that by equating corresponding coefficients. I hope that clarifies it. – Kaster Oct 23 '17 at 17:32

Let $f(x) = x^2$ and $g(x) = -x^2 + 2x - 2 \implies \dfrac{d}{dx}(f(x))|_{x = a} = 2a$ and $\dfrac{d}{dx}(g(x))|_{x = b} = -2b + 2$

$2a = -2b + 2 \implies a = 1 - b \implies$

$A: (1 - b, (1 - b)^2), B: (b, -b^2 + 2b - 2) \implies m_{AB} = \dfrac{2b^2 - 4b + 3}{1 - 2b} = -2b + 2 \implies$ $2b^2 - 2b - 1 = 0 \implies b = \dfrac{1 \pm \sqrt{3}}{2}$

Using $b = \dfrac{1 + \sqrt{3}}{2} \implies$

$A: (\dfrac{1 - \sqrt{3}}{2},\dfrac{2 - \sqrt{3}}{2}), B: (\dfrac{1 + \sqrt{3}}{2},\dfrac{\sqrt{3} - 4}{2}) \implies m_{AB} = 1 - \sqrt{3} \implies$ $\boxed{y = (1 - \sqrt{3})x + \dfrac{\sqrt{3} - 2}{2}}$

Using $b = \dfrac{1 - \sqrt{3}}{2} \implies A^{'}: (\dfrac{1 + \sqrt{3}}{2}, \dfrac{2 + \sqrt{3}}{2}), B^{'}: (\dfrac{1 - \sqrt{3}}{2}, \dfrac{-\sqrt{3} - 4}{2}) \implies$ $m_{A^{'}B^{'}} = 1 + \sqrt{3} \implies \boxed{y = (1 + \sqrt{3})x - \dfrac{\sqrt{3} + 2 }{2}}.$