Find the equation of a parabola and a point on the parabola given a two tangent lines.

Question

I've been trying to find a solution to this problem but I'm not too sure how to go about solving it.

I need to find the unknown values of $A, B$ and $C$ in the parabola equation $y(x) = Ax^2 + Bx + C$ given that the parabola passes through point (0,0) and is tangent to the line $y1(x) = 0.1x$ which also passes through the point (0,0).

I also need to find an unknown point (x-coordinate, y-coordinate) on the parabola which is tangent to $y2(x) = -0.08x + 10$ given that $y2(x) = -0.08x + 10$ passes through (200,-6).

I would greatly appreciate any help with solving this problem. So far I've tried to use the $y1(x)$ line to determine the vale of $B$ and $C$ in the parabola:

As the parabola passes through (0,0) I tried,

$$y(0) = 0$$ so $$A(0)^2 + B(0) + C = 0\\ C = 0$$

As the line $y1(x) = 0.1x$ is tangent to the parabola at (0,0) I tried,

$$ y'(0) = 0.1$$ so
$$ 2A(0) + B = 0.1\\ B = 0.1$$

However when trying to find a point on the parabola where the parabola is tangent to $y2(x)$ it seems to me that the value for B would be different, so this has me really confused on how to solve this problem.

Once again, any help with this would be very very appreciated :)

Answer 1

Hint: Let the tangent meet the parabola at $(x_0,y_0)=(x_0,Ax_0^2+0.1x_0)$. The slope of the tangent is $$-0.08 = 2Ax_0+0.1, \tag 1$$ and the equation of the line implies $$Ax_0^2+0.1x_0 = -0.08x_0+10. \tag 2$$ Solve them simultaneously to compute $x_0$ and $A$.

Answer 2

Your condition means that $y=0.1x$ is the linear approximation of your quadratic polynomial at $x=0$. Thus the "error" made by replacing the one by the other is quadratic.

Otherwise said, you are looking for an equation of the form:

$$y=0.1x + \underbrace{A x^2}_{\text{error term}}$$

with a general parameter $A$, with $B=0.1$ and $C=0$.

It is difficult for me to understand the second part of the problem. Could you give more precision ?

Answer 3

Since it passes through $(0,0)$ then $C=0$.

Since it has a tangent line on x=0, we know that its derivate must match the slope of the line it is tangent to.

So we have $y(x) = ax^2 + bx$

$y'(x) = 2ax + b$

$y'(0) = b = 0.1$

So know we can write:

$y(x) = ax^2+0.1x$

We are still missing the value of $a$ we can use the next chunk of data to see if it resolves or not.

According the next piece of information the parabola needs to intercept somewhere $(x_0,y_0)$ the tangent $y_2(x)$ this means that:

$y_0=ax_0^2+0.1x_0$ Because the line needs to touch we know: $y_2(x_0)=y_0 = ax_0^2+0.1x_0 = -0.08x_0 +10$

This allows us to calculate the missing a based on the value of $x_0$. I however suspect the problem is not well written and probably they are giving you the tangent point as well when they mention $(200,6)$.

In this case $x_0= 200$ and $y_0=6$

$a200^2+0.1\cdot200 = 6$

Which makes:

$a=\frac{-7}{100^2}$