Let $a$ and $b$ be real numbers with $a \neq b$. Let $z = x + iy$ be a complex number such that $$|z - a|^2 - |z - b|^2 = 1.$$ (a) (i) Prove that $\bar{x} = \frac{a + b}{2} + \frac{1}{2(b - a)}$. (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 $AE \perp BC$. The line $ED$ meets the line $BA$ at the point $F$. The point $G$ lies on the line $CD$ such that $FG \perp BC$. (i) Prove that $FADG$ is a cyclic quadrilateral. (ii) Explain why $\angle GFD = \angle ZED$. (iii) Prove that $GA$ is a tangent to the circle through the points $A, B, C$ and $D$. (c) A mass is attached to a spring and moves in a resistive medium. The motion of the mass satisfies the differential equation $$\frac{d^2y}{dt^2} + \frac{dy}{dt} + 2y = 0,$$ 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 $$y = Ae^{-2t} + Be^{-t}.$$ When $t = 0$, it is given that $y = 0$ and $\frac{dy}{dt} = 1$. Find the values of $A$ and $B$.

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Let $a$ and $b$ be real numbers with $a \neq b$ - HSC - SSCE Mathematics Extension 2 - Question 4 - 2011 - Paper 1

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Let $a$ and $b$ be real numbers with $a \neq b$. Let $z = x + iy$ be a complex number such that $$|z - a|^2 - |z - b|^2 = 1.$$ (a) (i) Prove that $\bar{x} = \fra... show full transcript

Worked Solution & Example Answer:Let $a$ and $b$ be real numbers with $a \neq b$ - HSC - SSCE Mathematics Extension 2 - Question 4 - 2011 - Paper 1

Step 1

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

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Answer

To prove this, start from the given equation: $|z - a|^2 - |z - b|^2 = 1.$ We can expand this:
$(x - a)^2 + (y - 0)^2 - ((x - b)^2 + (y - 0)^2) = 1.$
This simplifies to
$(x^2 - 2ax + a^2) - (x^2 - 2bx + b^2) = 1.$
This reduces to
$-2ax + a^2 + 2bx - b^2 = 1.$
Then we can combine the terms:
$2(b - a)x + (a^2 - b^2) = 1.$
Solving for $x$ , we get
$x = \frac{1 - (a^2 - b^2)}{2(b - a)} = \frac{a + b}{2} + \frac{1}{2(b - a)}.$

Step 2

Describe the locus of all complex numbers $z$ such that $|z - a|^2 - |z - b|^2 = 1$

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Answer

The equation leads to a locus described as a vertical line at:
$x = \frac{a + b}{2} + \frac{1}{2(b - a)}.$
This indicates that the locus does not represent a conic section like an ellipse, hyperbola, or parabola.

Step 3

Prove that $FADG$ is a cyclic quadrilateral

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Answer

To prove that $FADG$ is cyclic, we utilize the properties of angles subtended. We can show that the opposite angles of quadrilateral $FADG$ add up to 180 degrees. This follows from the angle of elevation and the inscribed angle theorem.

Step 4

Explain why $\angle GFD = \angle ZED$

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Answer

Since both angles subtend the same arc $WD$ on the circle, they are equal by the inscribed angle theorem.

Step 5

Prove that $GA$ is a tangent to the circle through the points $A, B, C$ and $D$

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Answer

To prove this, we consider the intersection of line $GA$ with the circle. Using the radius at point $A$ , if the radius at point $A$ is perpendicular to $GA$ , it confirms that $GA$ functions as a tangent to the circle.

Step 6

Show that $y = Af(t) + Bg(t)$ is a solution.

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Answer

Given that both $y=f(t)$ and $y=g(t)$ satisfy the differential equation, we substitute $y = Af(t) + Bg(t)$ and show the left-hand side equals zero, confirming it satisfies the equation.

Step 7

Show that only possible values of $k$ are $k = -1$ and $k = -2$

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Answer

Assuming $y = e^{kt}$ , substituting into the differential equation leads to the characteristic equation. Factoring results in $(k + 1)(k + 2) = 0$ , thus yielding $k = -1$ and $k = -2$ as the only solutions.

Step 8

Find the values of $A$ and $B$

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Answer

Using the initial conditions $y(0) = 0$ and $rac{dy}{dt} = 1$ when $t=0$ , we set up the equations:

$A + B = 0$
$-2A - B = 1$
From the first equation, we get $B = -A$ . Substituting this into the second gives $-2A + A = 1$ , thus $-A = 1$ , and we find $A = -1$ and $B = 1$ .

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Questions answered

Let $a$ and $b$ be real numbers with $a \neq b$ - HSC - SSCE Mathematics Extension 2 - Question 4 - 2011 - Paper 1

Worked Solution & Example Answer:Let $a$ and $b$ be real numbers with $a \neq b$ - HSC - SSCE Mathematics Extension 2 - Question 4 - 2011 - Paper 1

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

Describe the locus of all complex numbers $z$ such that $|z - a|^2 - |z - b|^2 = 1$

Prove that $FADG$ is a cyclic quadrilateral

Explain why $\angle GFD = \angle ZED$

Prove that $GA$ is a tangent to the circle through the points $A, B, C$ and $D$

Show that $y = Af(t) + Bg(t)$ is a solution.

Show that only possible values of $k$ are $k = -1$ and $k = -2$

Find the values of $A$ and $B$

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