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I have a problem that I need to solve for a current project. I need to find a quadratic equation which has only one root that it at (0,0) and there is one point know at any given time on the parabola that is at (10,60).

Please help me find a way to solve that. Thank you!

4 Answers 4

1

I assume you want a quadratic function of a single variable.

If the only root is at $(0,0)$, then the quadratic function is necessarily of the form $$f(x) = ax^2$$ and since you require $f(10)=60$, you must have $$60 = a(10)^2$$ $$100a = 60$$ $$a = \frac35$$ Thus $$\boxed{f(x) = \frac35 x^2}$$

0

Let $f(x) = ax^2 + bx + c$ for some $a,b,c \in \mathbb{R}$ be the wanted parabola. Then

  • one root at $(0, 0) \rightarrow f(0) = 0 = a \cdot 0^2 + b \cdot 0 + c \rightarrow c = 0$
  • $f(10) = 60 = a \cdot 10^2 + b \cdot 10 = 100a + 10b \rightarrow 6 = 10 \cdot a + b \rightarrow b = 6 - 10a$
  • only one root at $(0, 0) \rightarrow \Delta(0) = 0 = b^2-4ac = (6 - 10a)^2 - 4a\cdot 0 = (6 - 10a)^2 = 0 \iff 6 - 10a = 0 \iff a = \frac{3}{5 }$

so all the parabolae in the form $$f(x) = \frac{3}{5} \cdot x^2$$ satisfy the requirements.

  • 1
    This function will have a second root at $x=10 -\frac6a$ (if $a\neq 0$). I think OP wants the only root to be at $x=0$, doesn't he? – 2017-02-20
  • 0
    yup! sorry! updated – 2017-02-20
0

Case $1$: Of type $x^2=4ay $. If $(10,60) $ passes through this parabola, then we get, $$100=4a (60)\Rightarrow 4a = \frac {5}{3} $$ The parabola is: $$x^2 =\frac {5}{3} y $$

Case $2$: Of type $y^2=4ax $. If $(10,60) $ passes through this parabola, then we get, $$3600=4a (10) \Rightarrow 4a =360$$ The parabola is: $$y^2=360x $$

Hope it helps.

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$y=ax^2$ if $x=10$ then $y=60$ $60=a\cdot 10^2$ from there $a=3/5$ desired quadratic function is $y=\frac{3}{5}x^2$

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    You got $\;a\;$ upwards... – 2017-02-20