Question:
The diagram shows part of a straight line graph drawn to represent the equation $y=\frac{ax^2+b}{cx}$, where $a$, $b$, and $c$ are integers. Given that the line passes through $(4,9)$ and has gradient $-\frac{1}{4}$, find
(i) the value of $\frac{y}{x}$ where the straight line cuts the horizontal axis, and
(ii) the value of $a$, of $b$ and of $c$.
Attempt:
(i)
$$Y=mX+c$$ Gradient, $m=-\frac{1}{4}$
At $(4,9)$, $$9=-\frac{1}{4}(4)+c$$ $$c=10$$
Therefore, $Y=-\frac{1}{4}X+10$
When $Y=0$,
$$0=-\frac{1}{4}(X)+10$$ $$X=40$$
Since $X=\frac{x}{y}$, therefore $\frac{y}{x}=\frac{1}{40}$.
Is my solution correct?
(ii)
$$y=\frac{ax^2+b}{cx}$$
Divide the equation by $y$,
$$1=\frac{ax}{cy}+\frac{b}{cxy}$$ $$\frac{b}{cxy}=1-\frac{ax}{cy}$$ $$\frac{1}{xy}=-\frac{a}{b}\frac{x}{y}+\frac{c}{b}$$
Since $Y=-\frac{1}{4}X+10$, $$-\frac{a}{b}=-\frac{1}{4}$$ $$\frac{c}{b}=10$$
After this step, I do not know how to continue. I have two equations, but three unknowns. How do I get a third equation so that I can solve for $a$, $b$, and $c$?
Any suggestions / help is greatly appreciated. Thank you!
The equation $y=\frac{ax^2+b}{cx}$ can be written as
$ax^2-cxy+b=0$.
As you mentioned, $Y=-\frac{1}{4}X+10$
or $x^2-40xy+4 = 0$, by replacing $Y = \frac{1}{xy}, X = \frac{x}{y}$.
So, $a = 1, b = 4, c = 40$.