Move parabola to make it tangent to lines

Question

could anyone help me with something very easy (I hope). I have two lines. For example:

$y = \frac { x + 4.5 } { 6 }$

(which is green line on the graph)

and

$y = 1.2 x$

(which is red line on the graph)

And I also have parabola like that:

$y = x - 5 ( 0.5 x - 0.65 ) ^ { 2 }$

I need to find coordinates to move parabola to make it tangent to both lines and with shape of parabola untouched, and lines can't be moved.

Are there any general rules to make such things. In the google I always find how to draw lines tangent to parabola and go through the point. But I can't figure it out how to make it in other direction

Graph now looks like that:

Answer 1

You can move the parabola left-right by adding an offset to $x$ and you can move it up-down adding an offset to $y$. Let's solve the problem in the more general case. The original parabola is $y=ax^2+bx+c$. The new equation for the parabola will be $$y=a(x-x_0)^2+b(x-x_0)+c+y_0$$ We will need the derivative with respect to $x$ for the tangents, so $$y'=2a(x-x_0)+b$$ Now the equations of the lines are $$y=m_1x+n_1\\y=m_2x+n_2$$ When the parabola is tangent to a line, it means that they intersect at one point, and that the slopes at that points are identical. So say we have the tangent point with line 1 at $x_1$ and with line 2 at $x_2$, so you get the following system of equations: $$\begin{align}a(x_1-x_0)^2+b(x_1-x_0)+c+y_0&=m_1x_1+n_1\\ a(x_2-x_0)^2+b(x_2-x_0)+c+y_0&=m_2x_2+n_2\\ 2a(x_1-x_0)+b&=m_1\\2a(x_2-x_0)+b&=m_2 \end{align}$$

You have a system with four equations, with four unknowns $(x_0,y_0,x_1,x_2)$.

Answer 2

You can shift the graph of any function horizontally by a distance of $\Delta x$ by replacing $x$ with $(x-\Delta x)$, and vertically by simply adding $\Delta y$ to its value. If the curve is given implicitly by $F(x,y)$, the latter corresponds to replacing $y$ by $(y-\Delta y)$. In the same vein as Andrei’s answer, then, the problem is to find these offsets for your parabola. In contrast to that answer, we’ll work with the original parabola.

First, construct tangents to the parabola that are parallel to the two given lines. If the parabola is given by an equation of the form $y=f(x)$ (i.e., it’s the graph of $f$), an easy way to do this is to find the points at which the derivative of $f$ is equal to the two slopes. This is a matter of solving two very simple equations for $x$ and plugging those values into $f$. Now, find the intersection point $P$ of the two lines you’ve constructed, and the intersection $Q$ of the original two lines. The offset that you need is then simply $(\Delta x,\Delta y) = Q-P$. If you work in homogeneous coordinates, these computations can be done directly, without having to solve any other equations.

For your problem, we have $f(x) = x-5(0.5x-0.65)^2$, so $f'(x) = 1-5(0.5x-0.65)$. The slope of the first line is $1/6$, for which $x=49/30$ (assuming that those decimal fractions in $f$ are exact) and $y=f(49/30)=269/180$. For the second line the slope is $1.2$, giving $x=61/50$ and $y=303/250$. I assume that you know how to construct equations of the tangents through these points and finish the computation, but I’ll go through it with homogeneous coordinates to illustrate that method.

The line through two points is given by their cross product, as is the intersection of two lines. Also, if a line has slope $m$, then the point at infinity $(1,m,0)$ lies on that line. So, our two constructed tangent lines are $$(49/30,269,180,1)\times(1,1/6,0)=(-1/6,1,-11/9),$$ i.e., $y=x/6+11/9$, and $$(61/50,303/250,1)\times(1,12/10,0)=(-6/5,1,63/250).$$ Their intersection is therefore $$\left(-\frac16,1,-\frac{11}9\right)\times\left(-\frac65,1,\frac{63}{250}\right) \approx (1.474,1.509,1.033),$$ which we dehomogenize by dividing through by the last coordinate, yielding $P\approx(1.427,1.460)$. Intersecting the target lines, we have $$\left(\frac16,-1,\frac{45}{60}\right)\times\left(\frac{12}{10},-1,0\right) \approx (0.750,0.900,1.033),$$ or $Q\approx (0.726,0.871)$, from which $\Delta x \approx -0.70$, $\Delta y\approx -0.59$. The shifted parabola is therefore (to two decimal places) $$y = f(x+0.70)-0.59 = (x+0.70)-5(0.5(x+0.70)-0.65)^2-0.59.$$