Equation of a parabola that passes thorught 2 point with know slopes

Question

I want to be able to solve for the equation of this parabola.

Known Points A(2,1) Slope @ A=1/2 B(7.25,2.5) Slope @ B=1/5

nothing else is known/given, The picture shows that parabola's Axis of symmetry is the X Axis but this is not necessarily true.

Is this sufficient information?

I just though of something.

if you plot parallel lines that pass though the points given and mirror them around a line perpendicular to the slope at respective points the will intersect at the focus which opens up much more information!

Answer 1

You could try starting from a generic parabola: $f(x)=ax^2+bx+c$ whose derivative is $f^{\prime}(x)=2ax+b$. What you know is that $f(2)=1$ (this ensures that the point $(2,1)$ is on the graph), $f^{\prime}(2)=1/2$ (this means the slope at $x=2$ is $1/2$), $f(7.25)=2.5$ and $f^{\prime}(7.25)=1/5$. If you plug in those values you get a system of equations to solve:

$a(2)^{2}+b(2)+c=1$

$a(7.25)^{2}+b(7.25)+c=2.5$

$2a(2)+b=1/2$

$2a(7.25)+b=1/5$

If the system of equations has a solution (in terms of $a$, $b$, and $c$) then you have your parabola. If the system is inconsistent, then no such parabola exists. If the system has more than one solution, then the parabola is not unique.

Answer 2

$$y=ax^2+bx+c$$ $$y'=2ax+b$$ at $x=2$ the $y'=0.5$ $$0.5=2a(2)+b$$ at $x=7.25$ $y'=1/5$ $$1/5=2a(7.25)+b$$ solve the two equation to get $$a=-\frac{1}{35}$$ $$b=\frac{43}{70}$$

to find the $c$, we will use the main function

at $x=2:y=1$ $$1=-\frac{1}{35}(2)^2+\frac{43}{70}(2)+c$$ $$c=-\frac{4}{35}$$ but if you use the second point to find $c$, the value is not same

I think that there is a problem in the question

Answer 3

The axis of symmetry is indeed the $x$-axis. We note that $1/2 = 1/(2 \cdot 1)$ and $1/5 = 1/(2 \cdot 2.5)$; this yields the guess

$$ x = y^2+1 $$

which satisfies each of the conditions.

ETA: More generally, if we disregard the hint about the axis of symmetry, we consider the general form of the parabola (with some parameters adjusted from the usual form for the sake of convenience):

$$ (Ax+Cy)^2 + Dx + Ey + F = 0 $$

Plugging in the two points $(2, 1)$ and $(7.25, 2.5)$ gives us

$$ (2A+C)^2 + 2D + E + F = 0 $$ $$ (7.25A+2.5C)^2 + 7.25D + 2.5E + F = 0 $$

Implicit differentiation gives us

$$ 2(Ax+Cy)\left(A+C\frac{dy}{dx}\right) + D + E\frac{dy}{dx} = 0 $$

or

$$ \frac{dy}{dx} = - \frac{2A(Ax+Cy)+D}{2C(Ax+Cy)+E} $$

which produces

$$ 4AC+2C^2+E = -8A^2-4AC-2D $$ $$ 14.5AC+5C^2+E = -72.5A^2-25AC-5D $$

We therefore have four equations in five unknowns. Although it's easy to see that $A = E = 0, C = F = 1, D = -1$ is a solution, there may be others if we ignore the hint about the axis of symmetry.

We consider two cases.

Case I. Suppose $A = 0$. Then, without loss of generality, $C = 1$, and our equations are

$$ 1+2D+E+F = 0 $$ $$ 6.25+7.25D+2.5E+F = 0 $$ $$ 2+E = -2D $$ $$ 5+E = -5D $$

Clearly, $D = -1, E = 0, F = 1$, and $y^2-x+1 = 0$ is the only solution.

Case II. Let $A \not= 0$, then without loss of generality, $A = 1$, and we have

$$ (2+C)^2+2D+E+F = 0 $$ $$ (7.25+2.5C)^2+7.25D+2.5E+F = 0 $$ $$ 2C^2+8C+2D+E+8 = 0 $$ $$ 5C^2+39.5C+5D+E+72.5 = 0 $$

which has the solution $C = -3.5, D = 3, E = -10.5, F = 2.25$ when $A = 1$, or in integers, we have $(2x-7y)^2+12x-42y+9 = 0$ as another solution. However, this can be rewritten as $(2x-7y)^2+6(2x-7y)+9 = 0$, or $2x-7y = -3$. This is a degenerate parabola—a line (the implicit form of the derivative yields an indeterminate $0/0$)—so $x = y^2+1$ is indeed the only solution.