A Level Maths | 2 Past Exam Questions
A Level Maths | 2 Past Exam Questions- Introduction
In this particular blog post we are going to look at 2 A Level Maths past examination questions that involve circle geometry and straight line geometry.
This particular question is a longer question and would generally have been seen within section B of the old Core 1 maths exams or section B of the current AS maths exams that are taken at the end of year 12.
There are three parts to the question and we will examine the methods for each part as well as to provide an explanation and solution for each.
Circle Geometry - The Question
Circle Geometry – Part (i)
To answer this part of the question you first need to find the gradient of the line AC which is the radius. Once this has been done it is important to realise that the tangent is perpendicular to this rule, so the gradient of the tangent will be the negative reciprocal.
The equation of a straight line can then be used to determine the equation of the tangent. This is shown below:
Gradient of the line is given by \frac{3-7}{1-3}=\frac{-4}{-2}=2
Perpendicular gradient is -\frac{1}{2}
\begin{aligned} & y-y_1=m\left(x-x_1\right) \\ & y-7=-\frac{1}{2}(x-3) \\ & 2 \times(y-7)=-(x-3) \\ & 2 y-14=-x+3 \\ & 2 y=-x+17 \\ & x+2 y=17 \end{aligned}
You are given another equation of a straight line so in order to find the coordinates of the point T then you need to perform simultaneous equations using linear equations which should be straightforward. The solution is shown here:
\begin{aligned} & x+2 y=17 \\ & y=2 x-9 \\ & x+2(2 x-9)=17 \\ & x+4 x-18=17 \\ & 5 x=35 \end{aligned}\begin{aligned} & x=7 \\ & y=2 \times 7-9 \\ & =5 \end{aligned}Coordinates are (7,5)
You can choose the method of substitution or the method of elimination, it does not matter. Above you will see that a method of substitution has been used because the second equation is of the form “y = “.
Circle Geometry – Final Part
For this part of the question you again need to perform a simultaneous equations calculation. Because you are dealing with the equation of a circle which is a quadratic you have to perform the method of substitution.
If a straight line is a tangent to a circle or curve then you will obtain a repeated root and this is what you need to obtain in your working in order to show that there is a tangent. The working is shown below:
\begin{aligned} & \quad(x-1)^2+(y-3)^2=20 \\ & (x-1)^2+(2 x-9-3)^2=20 \\ & (x-1)^2+(2 x-12)^2=20 \\ & x^2-2 x+1+4 x^2-48 x+144=20 \\ & 5 x^2-50 x+125=0 \\ & x^2-10 x+25=0 \\ & (x-5)(x-5)=0 \\ & x=5 \& 5 \end{aligned}y=2 x-9 \text { is a tangent to the circle }y=2 \times 5-9=1 giving coordinates (5, 1)
You can see that a repeated root has been obtained, which is the condition for a tangent to exist.
If you have not obtained a repeated root then that is an indication that an error has taken place in your working and you need to go over what you have done once again.
A Level Maths - The Second Question
Part (i) of the question – the notation that is provided is asking you to complete the square. Because x^2 is not a unit then you need to take a factor of 4 out. The solution for part (i) is shown here:
\begin{aligned} & 4 x^2-24 x+27 \\ = & 4\left(x^2-6 x\right)+27 \\ = & 4\left\{(x-3)^2-3^2\right\}+27 \\ = & 4\left\{(x-3)^2-9\right\}+27 \\ = & 4(x-3)^2-36+27 \\ = & 4(x-3)^2-9 \end{aligned}
For part (ii) you are required to use the answer to part (i) to help. The solution is shown here:
Minimum point is (3,-9)
Once a quadratic is in the correct form it is very easy to determine the line of symmetry and also any turning points. There should be no further calculations performed.
For part (iii) again you should use the answer that has been obtained in part (i). Whilst you could use the quadratic formula, using the completed square form of answer is sufficient to help you to proceed in terms of obtaining a solution as shown here:
Part iii)
\begin{aligned} & 4 x^2-24 x+27=0 \\ & 4(x-3)^2-9=0 \\ & 4(x-3)^2=9 \\ & (x-3)^2=9 / 4 \\ & x-3= \pm \sqrt{\frac{9}{4}}= \pm \frac{3}{2} \end{aligned}\begin{aligned} x & =3+3 / 2 \text { or } 3-3 / 2 \\ & =4.5 \text { or } 1.5 \end{aligned}Part (iv) – here you simply need to use the answer in part (i) – using the turning point and where the curve crosses the y-axis. There are the points that are required to be labelled on any sketch as shown here:
