# Find quadratic equation that has only one root at (0,0) and passes through (10,60) [closed]

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).

## closed as off-topic by Antonios-Alexandros Robotis, kingW3, TheGeekGreek, projectilemotion, NamasteFeb 20 '17 at 21:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Antonios-Alexandros Robotis, kingW3, TheGeekGreek, projectilemotion, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.

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 $$\boxed{f(x) = \frac35 x^2}$$

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.

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.

• 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? – MPW Feb 20 '17 at 13:47
• yup! sorry! updated – sirfoga Feb 20 '17 at 13:51

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

• You got $\;a\;$ upwards... – DonAntonio Feb 20 '17 at 13:50