Let $a$ and $b$ be real numbers with $a \neq b$ - HSC - SSCE Mathematics Extension 2 - Question 4 - 2011 - Paper 1
Question 4
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
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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 (x2−2ax+a2)−(x2−2bx+b2)=1.
This reduces to −2ax+a2+2bx−b2=1.
Then we can combine the terms: 2(b−a)x+(a2−b2)=1.
Solving for x, we get x=2(b−a)1−(a2−b2)=2a+b+2(b−a)1.
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=2a+b+2(b−a)1.
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=ekt, 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.
