2010 Paper 1 Q12 - equation of circles

Today is Deepavali. Best wishes to our Hindu friends!

We are now down to the last question of paper 1. So far, I am sure that most of you will agree that the paper is rather easy. Which is of course not good news too. This is because if many people found it easy, getting an A1 would not be so easy…

Anyway, let’s look at this last question

(i) Write down the equation of the circle with centre A(-3 , 2) and radius 5.

This circle intersects the y-axis at points P and Q.

(ii) Find the length of PQ.

A second circle, centre B, also passes through P and Q.

(iii) State the y-coordinate of B.

Given that the x-coordinate is positive and that the radius of the circle is √80, find

(iv) the x-coordinate of B

The equation of the circle, centre B which passes through P and Q may be written in the form

x² + y² +2gx + 2fy + c = 0.

(v)  State the value of g and f and find the value c.

Answer:

(i)  The first question is a real slam-dunk question. It really tests your memory of how the equation of a circle is.

So, let us recap. The equation of a circle, centre (h,k) and radius r, is given by

(x – h)² + (y – k)² = r²

Hence, the equation of the circle is  (x + 3)² + (y – 2)² = 5²

so,          x² + 6x + 9 + y² – 4y + 4 = 25

                x² + y² + 6x – 4y - 25 + 13 = 0

                x² + y² + 6x – 4y – 12 = 0

(ii) At the y-axis, x = 0. Therefore,   y² – 4y – 12 = 0

                                                                (y -  6)(y + 2) = 0

                                                                y = 6 or y = -2

Hence, P (0 , 6) and Q(0 , -2)  and the length PQ = 8 units.

(iii)  The y coordinate of B will be the midpoint of PQ and also the same y-coordinate of the first circle. Hence, the y-coordinate of B is 2.

(iv) For part (iv), we need to remember that the radius of the circle of centre B is the distance of B to any point in the circle. Since the radius of the second circle is √80, then, it means that PB = √80.

Let B be (p , q), then

PB² = p² + (6 – 2)² = 80

p² + 16 = 80

p² = 64

p = ±8 

But given that the x axis is positive, then the x-coordinate of B is 8.

(v)  Now that we have the coordinates of B(8, 2) and the radius of the circle, √80, we can put them into the equation of a circle as such:

(x – 8)² + (y – 2)² = 80

x² – 16x + 64 + y² – 4y + 4 – 80 = 0

x² + y² – 16x – 4y – 12 = 0

Hence, g = -8, f = -2 (which is really –h and –k respectively) and c = -12, which is really –(r² – h² – k²)

2010 Paper 1 Q11 - tangents and normals

A curve is such that  

 

(i) Given that the curve passes through the point (1, 5), find the equation of the curve

(ii) Find the x-coordinates of the stationary points of the curve

(iii) Obtain an expression for d²y/dx² and hence, or otherwise, determine the nature of each stationary point.

(i) We all know that to get the equation of the curve from the differential, we need to integrate the differential. But don’t forget the constant of integration.

At (1, 5), then

5 = -8 – 2 + c

c = 15

Therefore the equation of the curve is:

(ii) At the stationary points, dy/dx = 0

Therefore,

Hence, the x-coordinates of the stationary points are 2 and -2

(iii) 

Hence, at x = 2, d²y/dx² is negative, Þ it is a max point

and when x = -2, it is a min point.

And that's all for today....

Tomorrow, we will do the last question of the paper 1 and move into paper 2.

2010 Paper 1 Q10 - trigonometric identities

Without using a calculator, show that

(i) tan 75° = 2 + √3,

(ii) sec² 75° = 4 tan 75°

Haha…wicked! I love these questions. For some, it would be a pain…why cannot use the calculator? Afterall, that’s what the calculator is for, isn’t it? Well, anyway, it is not the answer that is important, it is the thought process. This is deceptively simple. Let’s get at it…

Now, when we think about this, there is not much into it that we can do. 75° is not a special angle. So how? Well, 75° may not be a special angle but it is the sum of two special angles: 30° and 45°! So, now we can translate

tan 75° = tan (30° + 45°). So now, we will use the tan (A + B) rule.

Now, tan 30° = 1/√3   and  tan 45° = 1

So, 

 Rationalising the denominator, we get,

2010 Paper 1 Q9 - roots of quadratic equation

Q:  Given that the roots of 3x² – 2x + 1 = 0 are α and β, find the quadratic equation whose roots are α+2β and 2α+β

Ahhh…roots of the equation… I love quadratic!

OK, to answer this type of question, we must first recall that ANY quadratic equation can be defined by the sum and products of its roots as such:

                x² – (sum of roots)x + product of roots = 0

This is a GIVEN. One must always notice that the quadratic equation is given by x² and not ax².  Hence, in this question, given that the quadratic equation begins with 3x², it means that we have to divide throughout by 3 first. This gives us

Hence,

α+β = 2/3   and   αβ = 1/3

If there is another quadratic equation with roots α+2β and 2α+β, then, the sum of the roots are

α+2β + 2α+β = 3α+3β = 3(α+β) = 2

The product of the roots will be (α+2β)(2α+β)

                                                = 2α² + 5αβ + 2β²

                                                = 2(α+β)² + αβ

                                                = 2(4/9) + 1/3

                                                = 8/9 + 3/9

                                                = 11/9

Hence, the quadratic equation with roots of α+2β and 2α+β is

2010 Paper 1 Q8 - tangents

The equation of a curve is y = x³ + 3x² – 9x + k, where k is a constant.

(i) Find the set of values of x for which y is decreasing.

(ii) Find the possible values of k for which the x-axis is a tangent to the curve.

Hmm….it seems that there are quite a bit of differential questions in this paper.  I guess that is good. We may be tending towards the earlier years when I was taking the A maths when more than 50% was calculus.

Anyway, let’s discuss part (i). When is it that the function is increasing, or when the function is decreasing?

Well, we return to dy/dx!  Again, what does dy/dx mean? It means the change in y with respect to a change in x. If a change in x creates a positive change in y, it means that y is increasing with every change in x.   When a change in x leads to a negative change in y, that means that y is decreasing.  Hence, when we want to find the set of values of x for which y  is decreasing, then we are going to find when dy/dx < 0.

Given that y = x³ + 3x² – 9x + k, then

For decreasing values of y, then let dy/dx < 0

Hence the solution set is  {x: x Є R,  - 3 < x < 1 }

Again, don't forget to put it in set notation because the question asked for the solution set.

(ii) To find the possible values of k, (and in this case we know that there will be more than one value of k since the question asked for possible values) , we need to remember that k is a constant. In this case, the only effect that k has in the curve is to move it up or down. Hence, what we want to know is, which value of x will produce a gradient of zero (parallel to the x-axis) and then from there, put in the value of x into the equation for y, and then equating it to zero, and we will find the value of k. Confused? Nevermind. Let’s see it in action. That is clearer.

When dy/dx = 0, then  x = 1 or x = -3

Substitute x = 1 into y  and equating that to zero brings:

1³ + 3(1)² – 9(1) + k = 0

1 + 3 – 9 + k = 0

k = 5

Substitute x = -3 into y and equating that to zero gives:

(-3)³ + 3(-3)² – 9(-3) + k = 0

-27 + 27 + 27 + k = 0

k = -27

Hence, when k = 5 or k = -27, the x-axis is a tangent to the curve.