SPM Practice 2 (Question 11 & 12)

Question 11 (3 marks):
Diagram 6 shows the graph of a straight line $\frac{x^2}{y}\text{ against }\frac{1}{x}.$  

Diagram 11

Based on Diagram 6, express y in terms of x.


Solution:

$\begin{array}{l}m=\frac{4-\left(-5\right)}{6-0}=\frac{3}{2}\\ c=-5\\ Y=\frac{x^2}{y}\text{, }X=\frac{1}{x}\\ \\ Y=mX+c\\ \frac{x^2}{y}=\frac{3}{2}\left(\frac{1}{x}\right)+\left(-5\right)\\ \frac{x^2}{y}=\frac{3}{2x}-5\\ \frac{x^2}{y}=\frac{3-10x}{2x}\\ \frac{y}{x^2}=\frac{2x}{3-10x}\\ y=\frac{2x^3}{3-10x}\end{array}$

Question 12 (3 marks):
The variables x and y are related by the equation $y=x+\frac{r}{x^2}$ , where r is a constant. Diagram 8 shows a straight line graph obtained by plotting $\left(y-x\right)\text{ against }\frac{1}{x^2}.$

Diagram 12

Express h in terms of p and r.


Solution:

$\begin{array}{l}y=x+\frac{r}{x^2}\\ y-x=r\left(\frac{1}{x^2}\right)+0\\ Y=mX+c\\ m=r,c=0\\ \\ m=\frac{y_2-y_1}{x_2-x_1}\\ r=\frac{5p-0}{\frac{h}{2}-0}\\ \frac{hr}{2}=5p\\ hr=10p\\ h=\frac{10p}{r}\end{array}$