Question 4 (15 marks) Use a SEPARATE writing booklet. (a) Let a and b be real numbers with a ≠ b. Let z = x + iy be a complex number such that |z − a|² − |z − b|² = 1. (i) Prove that x = \frac{a+b}{2} + \frac{1}{2(b-a)}. (ii) Hence, describe the locus of all complex numbers z such that |z − a|² − |z − b|² = 1. (b) In the diagram, ABCD is a cyclic quadrilateral. The point E lies on the circle through the points A, B, C and D such that AE || BC. The line ED meets the line BA at the point F. The point G lies on the line CD such that FG || BC. Copy or trace the diagram into your writing booklet. (i) Prove that FADG is a cyclic quadrilateral. (ii) Explain why ∠GFD = ∠AED. (iii) Prove that GA is a tangent to the circle through the points A, B, C and D. Question 4 (continued) (c) A mass is attached to a spring and moves in a resistive medium. The motion of the mass satisfies the differential equation d^2y/dt^2 + dy/dt + 2y = 0, where y is the displacement of the mass at time t. (i) Show that, if y = f(t) and y = g(t) are both solutions to the differential equation and A and B are constants, then y = Af(t) + Bg(t) is also a solution. (ii) A solution of the differential equation is given by y = e^(kt) for some values of k, where k is a constant. Show that the only possible values of k are k = −1 and k = −2. (iii) A solution of the differential equation is given by y = Ae^(-2t) + Be^(-t). When t = 0, it is given that y = 0 and dy/dt = 1. Find the values of A and B.

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Question 4 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 4 - 2011 - Paper 1

Question 4

Question 4 (15 marks) Use a SEPARATE writing booklet. (a) Let a and b be real numbers with a ≠ b. Let z = x + iy be a complex number such that |z − a|² − |z − b|² ... show full transcript

Worked Solution & Example Answer:Question 4 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 4 - 2011 - Paper 1

Step 1

a) (i) Prove that x = \frac{a+b}{2} + \frac{1}{2(b-a)}.

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Answer

To prove this, we start with the equation given:

$|z - a|^2 - |z - b|^2 = 1.$
This can be expressed as:
$|(x + iy) - a|^2 - |(x + iy) - b|^2 = 1$
Expanding gives:
$(x-a)^2 + y^2 - ((x-b)^2 + y^2) = 1.$
The $y^2$ terms cancel out, resulting in:
$(x-a)^2 - (x-b)^2 = 1.$
Factoring, we get:
$(x - a - (x - b))(x - a + (x - b)) = 1.$
This simplifies to:
$(b-a)(2x - (a + b)) = 1.$
Thus, solving for $x$ , we have:
$x = \frac{a + b}{2} + \frac{1}{2(b - a)}.$

Step 2

a) (ii) Hence, describe the locus of all complex numbers z such that |z − a|² − |z − b|² = 1.

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The locus of the complex numbers $z$ satisfying the equation
$|z - a|^2 - |z - b|^2 = 1$
represents a vertical line in the complex plane. This line can be derived from part (i), where the expression for $x$ yields a constant value. The resulting line is vertical and intersects the real axis at the point
$x = \frac{a+b}{2} + \frac{1}{2(b-a)}.$

Step 3

b) (i) Prove that FADG is a cyclic quadrilateral.

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A cyclic quadrilateral is one where all points lie on the circumference of a circle. To demonstrate that quadrilateral FADG is cyclic, we can show that opposite angles sum to 180 degrees.
Using the properties of cyclic quadrilaterals and the angles formed by the chords AD and FG, we state that ∠FAD + ∠FGD = 180° (since they subtend the same arc). Thus, we conclude that FADG is cyclic.

Step 4

b) (ii) Explain why ∠GFD = ∠AED.

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The angles ∠GFD and ∠AED are equal because they are angles subtended by the same chord (AD) in the cyclic quadrilateral formed by points A, B, C, D, and E. Therefore, by the properties of cyclic quadrilaterals, we can state that these angles are equal.

Step 5

b) (iii) Prove that GA is a tangent to the circle through the points A, B, C and D.

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To prove that GA is a tangent, we show that the angle between the tangent GA and the radius drawn to the point of tangency (point A) is 90 degrees. By the circle's properties, we note that angle ∠GAB must equal ∠EAD, which are subtended by arc AD. Therefore, from the cyclic nature of quadrilateral FADG, GA is indeed a tangent to the circle.

Step 6

c) (i) Show that y = Af(t) + Bg(t) is also a solution.

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Given that both functions y = f(t) and y = g(t) satisfy the differential equation
$\frac{d^2y}{dt^2} + \frac{dy}{dt} + 2y = 0,$
we substitute y with Af(t) + Bg(t). Taking the relevant derivatives, we find:
$y' = Af'(t) + Bg'(t)$
$y'' = Af''(t) + Bg''(t).$
Substituting these into the original equation leads to each term combining to show that
$Af''(t) + Bg''(t) + Af'(t) + Bg'(t) + 2(Af(t) + Bg(t)) = 0,$
confirming that y = Af(t) + Bg(t) is indeed a solution.

Step 7

c) (ii) Show that the only possible values of k are k = −1 and k = −2.

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Given the solution of the differential equation as y = e^(kt), we differentiate twice:
$y' = ke^{kt},$
$y'' = k^2e^{kt}.$
Substituting into the original equation yields:
$k^2e^{kt} + ke^{kt} + 2e^{kt} = 0.$
Factoring out e^(kt) gives us:
$(k^2 + k + 2) = 0.$
The roots of this equation can be calculated using the quadratic formula, leading to the possible values of k as k = -1 and k = -2.

Step 8

c) (iii) Find the values of A and B.

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Given the function y = Ae^(-2t) + Be^(-t) and the conditions that y(0) = 0 and dy/dt at t = 0 = 1, we substitute t = 0 into the equation:
$A + B = 0.$
Additionally, by computing the derivative:
$dy/dt = -2Ae^{-2t} - Be^{-t},$
setting t = 0 gives us:
$-2A - B = 1.$
We solve the simultaneous equations:

A + B = 0
-2A - B = 1.
Substituting B from equation 1 into equation 2 yields A = -1, hence B = 1. Therefore, the values are A = -1 and B = 1.

Question 4 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 4 - 2011 - Paper 1

Worked Solution & Example Answer:Question 4 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 4 - 2011 - Paper 1

a) (i) Prove that x = \frac{a+b}{2} + \frac{1}{2(b-a)}.

a) (ii) Hence, describe the locus of all complex numbers z such that |z − a|² − |z − b|² = 1.

b) (i) Prove that FADG is a cyclic quadrilateral.

b) (ii) Explain why ∠GFD = ∠AED.

b) (iii) Prove that GA is a tangent to the circle through the points A, B, C and D.

c) (i) Show that y = Af(t) + Bg(t) is also a solution.

c) (ii) Show that the only possible values of k are k = −1 and k = −2.

c) (iii) Find the values of A and B.

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