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Solution to 2008 Higher Level Mathematics Paper 2

The questions are worded differently, due to possible copyright issues

Section A

Question 1

(a)Given a circle that contains the point (1,3) and with centre (-3,2), find the equation of that circle.

The distance between the two points is the radius of the circle:

$|d|\ =\ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\ =\ \sqrt{(-3 - 1)^2 + (2 - 3)^2} = \sqrt{17}$

The general equation of a circle:

$(x-h)^2+(y-k)^2=r^2$

$(x+3)^2+(y-2)^2=17$

(b)(i)Given a circle with equation , prove that a tangent to this circle at  has the equation

This is one of the standard proofs on the Higher Mathematics Curriculum.

   (ii) The tangent to the circle  at the point (2,3), crosses the x-axis at (k,0). Find k.

By part (i), the equation of that line is:

$2x + 3y = 13$

Let $y=0$ :

$2x = 13$

$x = \frac{13}{2}$

(c) The line  contains the centre of a circle R. R also contains the points m(8,5) and n(9,-2).
   (i)Find the equation of R.

A line through the midpoint of m and n, perpendicular to the line joining m and n, will contain the centre of R.

$C\ =\ \left ( \frac{x_1 + x_2}{2}\ ,\ \frac{y_1 + y_2}{2} \right )\ =\ (\frac{17}{2}\ ,\ \frac{3}{2})$

$S\ =\ \frac{y_2 -y_1}{x_2 - x_1}\ =\ \frac{-7}{1}\ =\ -7$

$S^{\perp}\ =\ \frac{1}{7}$

The equation of a line through $C$ , with slope $S^{\perp}$ :

$y - y_1\ =\ m(x - x_1)$

$y - \frac{3}{2}\ =\ \frac{1}{7}(x - \frac{17}{2})$

$7y-x=2$

The point at which this line and P cross is the centre-point of the circle. Using simultaneous equation we find this point to be (5,1). The radius is the distance between m/n and (5,1):

$|d|\ =\ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\ =\ \sqrt{(8-5)^2 + (5-1)^2} = \sqrt{25}\ =\ 5$

Therefore the equation of R is:

$(x-5)^2 +(y-1)^2 = 25$

(ii) The point l is some point on the major arc of R. Prove that .

Let us first determine $\angle mon$ :

Slope of mo:

$S_{mo} \ =\ \frac{y_2 -y_1}{x_2 - x_1}\ =\ \frac{5-1}{8-5}\ =\ \frac{4}{3}$

Slope of no:

$S_{no} \ =\ \frac{y_2 -y_1}{x_2 - x_1}\ =\ \frac{-2-1}{9-5}\ =\ -\frac{3}{4}$

As expected, they are perpendicular, and therefore $\angle mon\ =\ 90^{\circ}$ .

Now, o is the centrepoint of the circle. Draw a line joining m and l, n and l, m and o and n and o. Also draw a line from l to o, extending sightly beyond o, call this point q. (1) The external angle $\angle moq$ is equal to the sum of the internal angles $\angle olm$ and $\angle oml$ and vice versa on the other side with $\angle noq$ . From above:

$90^{\circ}\ =\ \angle moq\ +\ \angle noq$

and by (1):

$90^{\circ}\ =\ \angle oln\ +\ \angle olm\ +\angle oml\ +\ \angle onl\$

Since mo and ol and no are all radii of the circle they are therefore of the same length. Therefore the angles opposite of them are equal:

$90^{\circ}\ =\ 2\angle oln\ +\ 2\angle olm\$

$45^{\circ}\ =\ \angle oln\ +\ \angle olm\$

$45^{\circ}\ =\ \angle mln$

Question 2

(a) . Find c, for c  R.

$|10\vec i + c\vec j|\ =\ \sqrt{10^2+c^2}\ =\ \sqrt{100+c^2}$

$|11\vec i - 2\vec j|\ =\ \sqrt{11^2+(-2)^2}\ =\ \sqrt{125}$

$\sqrt{100+c^2}\ =\ \sqrt{125}$

$100+c^2\ =\ 125$

$c\ =\ \pm5$

(b)  and 
   ...
   (i)Determine the value of x, if

Vector perpendicular to a:

$k(-3\vec i - \vec j)$

$k(-3\vec i - \vec j)\ =\ (-\vec i + 3\vec j) - x(4\vec i - 2\vec j)$

$-3k\vec i - k\vec j\ =\ -\vec i + 3\vec j - 4x\vec i + 2x\vec j$

$-3k\vec i - k\vec j\ =\ -(1+4x)\vec i + (3+2x)\vec j$

$-3k = -(1+4x),\ -k = (3+2x)$

$3(3+2x) = -(1+4x)$

$x=-1$

  (ii) If x is the origin, determine the value of

$\cos \theta\ =\ \frac{\vec a.\vec c}{|\vec a||\vec c|}\ = \frac{(-\vec i + 3\vec j).(4\vec i - 2\vec j)}{|(-\vec i + 3\vec j)||(4\vec i - 2\vec j)|}\ =\ \frac{-10}{\sqrt{200}}\ =\ -\frac{1}{\sqrt{2}}$

$\theta\ =\ 135^{\circ}$

(c) Given a parallelogram oabc, define d as the midpoint of [oa], and define p as the point at which [db] crosses the diagonal [ac].
   (i) Express  in terms of ,  and , given .

$\vec p\ =\ \vec {oa}\ +\ \vec {ap}$

$\vec p\ =\ \vec {a}\ +\ k\vec {ac}$

$\vec p\ =\ \vec {a}\ +\ k[\vec c\ -\ \vec a]$

$\vec p\ =\ \vec {a}\ +\ k\vec c\ -\ k\vec a$

$\vec p\ =\ (1-k)\vec {a}\ +\ k\vec c\$

   (ii) If  lR, express  in terms of ,  and l.

$\vec p\ =\ \vec {oa}\ +\ \vec {ab}\ +\ \vec {bp}$

$\vec p\ =\ \vec {a}\ +\ \vec {oc}\ +\ l\vec {bd}$

$\vec p\ =\ \vec {a}\ +\ \vec {c}\ +\ l[\vec d\ -\ \vec b]$

$\vec p\ =\ \vec {a}\ +\ \vec {c}\ +\ l\vec d\ -\ l\vec b$

$\vec p\ =\ \vec {a}\ +\ \vec {c}\ +\ \frac{1}{2}l\vec a \ -\ l[\vec {oa}\ +\ \vec {ab}]$

$\vec p\ =\ \vec {a}\ +\ \vec {c}\ +\ \frac{1}{2}l\vec a \ -\ l\vec {a}\ -\ l\vec {c}$

$\vec p\ =\ \vec {a}\ +\ \vec {c}\ -\ \frac{1}{2}l\vec a \ -\ l\vec c$

$\vec p\ =\ (1-\frac{l}{2})\vec {a}\ +\ (1-l)\vec {c}$

   (iii) Find the value of k and l.

$(1-k)\vec {a}\ +\ k\vec c\ =\ (1-\frac{l}{2})\vec {a}\ +\ (1-l)\vec {c}$

$(1-k)\ =\ (1-\frac{l}{2}),\ k\ =\ (1-l)$

$(1-(1-l))\ =\ 1-\frac{l}{2}$

$l\ =\ 1-\frac{l}{2}$

$l\ +\ \frac{l}{2}\ =\ 1$

$l =\ \frac{2}{3}$

$k =\ \frac{1}{3}$

Question 3

(a) ,  are parametric equations of a line. Express the line as one equation in x and y.

$x\ =\ 7t\ -\ 4$

$t\ =\ \frac{x+4}{7}$

$y\ =\ 3\ -\ 3t$

$t\ =\ \frac{3-y}{3}$

$\frac{x+4}{7}\ =\ \frac{3-y}{3}$

$3x\ +\ 12\ =\ 21\ -\ 7y$

$3x\ +\ 7y\ =\ 9$

(b)Define four points: a(2,1) b(10,7) c(14,10) d(7,1)
   (i) Plot the points on the coordinate plane.
   (ii) Determine if , and if

$|ab|\ =\ \sqrt{(10-2)^2 + (7-1)^2}\ =\ \sqrt{100}\ =\ 10$

$|bc|\ =\ \sqrt{(14-10)^2 + (10-7)^2}\ =\ \sqrt{25}\ =\ 5$

$|ab|\ =\ 2|bc|$

$|ad|\ =\ \sqrt{(7-2)^2 + (1-1)^2}\ =\ \sqrt{25}\ =\ 5$

$|ab|\ =\ 2|ad|$

   (iii) The transform  is defined , .
         ...
         Find , the images of a,b,c and d under the transform f.

$a(2,1) \rightarrow a^{'}((2+1),(2-2(1))) \rightarrow a^{'}(3,0)$

$b(10,7) \rightarrow b^{'}((10+7),(10-2(7))) \rightarrow b^{'}(17,-4)$

$c(14,10) \rightarrow c^{'}((14+10),(14-2(10))) \rightarrow c^{'}(24,-6)$

$d(7,1) \rightarrow d^{'}((7+1),(7-2(1))) \rightarrow d^{'}(8,5)$

Who Added These Notes?

Zorba

Solutions:Maths HL 2008 Paper 2

From ZuluNotes - Free Leaving Cert Notes

Contents

Section A

Question 1

Question 2

Question 3

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