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 :)

  • 1
    $\begingroup$ The fact that the second line passes through $(200,-6)$ is irrelevant. $\endgroup$ – amd Sep 19 '17 at 20:36

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 $A$.

  • $\begingroup$ Please correct me if I'm understanding this incorrectly. Equation (2) would become Ax^2 + 0.18x = 10, and so to solve both equations simultaneously you would need to subtract equation 2 from 1 or 1 from 2. However, equation 2 is in quadratic form while equation 1 is in linear form, so that wouldn't allow you to subtract one from the other. $\endgroup$ – Dani Green Sep 19 '17 at 20:47
  • $\begingroup$ Eq. (1) implies $Ax_0 = -0.09 \implies Ax_0^2 = -0.09 x_0$. Plug in this value in Eq. (2) to compute $x_0$. After computing $x_0$, use $A x_0 = -0.09$ to find the value of $A$. $\endgroup$ – Math Lover Sep 19 '17 at 20:49

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 ?

  • $\begingroup$ For the second part of the problem, the parabola equation is tangent to y2(x) = -0.08x + 10 at an unknown point, we are supposed to find this unknown point with the only information given is that that line (y2(x)) passes through the point (200,-6). $\endgroup$ – Dani Green Sep 19 '17 at 20:35
  • $\begingroup$ Thus, it is the parabola found in the first part ? $\endgroup$ – Jean Marie Sep 19 '17 at 20:38
  • $\begingroup$ Yeah, sorry should of mentioned that, the two problems relate to the same parabola. $\endgroup$ – Dani Green Sep 19 '17 at 20:41
  • $\begingroup$ @DaniGreen The fact that the line passes through $(200,-6)$ is irrelevant. $\endgroup$ – amd Sep 19 '17 at 20:45

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:


  • $\begingroup$ The point (200,-6) was not the tangent point, it was the point we had to use to calculate y2(x) given that the slope or gradient of y2(x) was -8% and passed through it. $\endgroup$ – Dani Green Sep 19 '17 at 21:40
  • $\begingroup$ I have solved the problem using Math Lovers suggested method, and the values I calculated for (xo, yo) are 1000/9 and 10/9 respectively. I have compared these results to the values on a graph of this problem, and the answers seem to match correctly. $\endgroup$ – Dani Green Sep 19 '17 at 21:45
  • $\begingroup$ Both of you are totally right $\endgroup$ – abr Sep 19 '17 at 21:51

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