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A family of curves, depending on parameters (A,B,C) has equation

$y(t)=A^{B/(B-1)}(A-(1-B)∗C∗t)^{1/(1-B)}$

I am looking for a curve of that family that passes through 3 known points. It looks like a non-linear system to solve.

If the 3 known points are M1(t1,y1=y(t1)), M2(t2,y2=y(t2)) and M3(t3,y3=y(t3)).

The system looks like

$y(t1)=A^{B/(B-1)}(A-(1-B)∗C∗t1)^{1/(1-B)}$

$y(t2)=A^{B/(B-1)}(A-(1-B)∗C∗t2)^{1/(1-B)}$

$y(t3)=A^{B/(B-1)}(A-(1-B)∗C∗t3)^{1/(1-B)}$

How can we solve it for A,B,C ? How many solutions exist ?

This question is closed to a previous one I asked, but the equation of family is different.

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    $\begingroup$ Are $b$ and $B$ the same parameter? $\endgroup$
    – Emilio Novati
    Jan 9, 2016 at 14:16
  • $\begingroup$ Yes. Thank you for pointing it. I fixed the typo. $\endgroup$
    – Philippe Morin
    Jan 9, 2016 at 15:10

1 Answer 1

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Surprizingly or not, this model is exactly the same as the previous one with slight changes of notations.

In $$y=a^{\frac{b}{b-1}} (a-(1-b) c x)^{\frac{1}{1-b}}$$ make $$a=\alpha \qquad c =\alpha \gamma \qquad b=1+\beta$$ and get $$y=\alpha (1+\beta \gamma x)^{-1/\beta }$$ So, the previous method works : just compute as before parameters $\alpha,\beta,\gamma$ and, at solution, go back to $a,b,c$ from the above definitions.

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