Eric Nicholas K's answer to lim's Junior College 2 H3 Maths Singapore question.
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First two parts.
Will look at the rest tonight, subject to my mental capacity for this chapter.
Will look at the rest tonight, subject to my mental capacity for this chapter.
Date Posted:
4 years ago
Second part to show is easy. The gradient from P to 1 + 0i is the same as the gradient from 1 + 0i to Q
Edit :
The two tangent values are equal.
The line from P(x,yi) to (1,0i) is parallel to the line from Z-1 (x - 1, yi) to the origin (0,0i). It's a horizontal translation of 1 unit to the left on the real(Re) axis.
So the angle Z-1 makes with the real axis and origin is equal to that made by P with the real axis and (1,0i)
The gradient of the 2 lines is the same
i.e tan [arg(z-1)] = gradient of PQ
Similarly,
The gradient of PQ equals gradient of the line from Q (4/z) to (1,0i).
Reflecting Q (4/z) along the line Re(z) = Im(z) gives -4/z.
Then, horizontal translation of 1 unit to the right on the real axis gives 1 - 4/z, which actually lies on the line from Z - 1 to (0,0i)
This means that 1 - 4/z is a reflection of 4/z along the real axis and tan[arg(1 - 4/z)] is equal to the gradient of line from Z-1 (x - 1, yi) to the origin (0,0i).
i.e tan[arg(1 - 4/z)] = tan [arg(z-1)]
So y/(x - 1) = 4y/(x² + y² - 4x)
y(x² + y² - 4x) = 4y(x - 1)
y(x² + y² - 4x - 4(x - 1)) = 0
y(x² - 8x + 4 + y²) = 0
y(x² - 2(4)(x) + (-4)² + 4 - (-4)² + y²) = 0
y( (x - 4)² + y² - 12) = 0
y = 0 or (x - 4)² + y² = 12
(Shown)
The two tangent values are equal.
The line from P(x,yi) to (1,0i) is parallel to the line from Z-1 (x - 1, yi) to the origin (0,0i). It's a horizontal translation of 1 unit to the left on the real(Re) axis.
So the angle Z-1 makes with the real axis and origin is equal to that made by P with the real axis and (1,0i)
The gradient of the 2 lines is the same
i.e tan [arg(z-1)] = gradient of PQ
Similarly,
The gradient of PQ equals gradient of the line from Q (4/z) to (1,0i).
Reflecting Q (4/z) along the line Re(z) = Im(z) gives -4/z.
Then, horizontal translation of 1 unit to the right on the real axis gives 1 - 4/z, which actually lies on the line from Z - 1 to (0,0i)
This means that 1 - 4/z is a reflection of 4/z along the real axis and tan[arg(1 - 4/z)] is equal to the gradient of line from Z-1 (x - 1, yi) to the origin (0,0i).
i.e tan[arg(1 - 4/z)] = tan [arg(z-1)]
So y/(x - 1) = 4y/(x² + y² - 4x)
y(x² + y² - 4x) = 4y(x - 1)
y(x² + y² - 4x - 4(x - 1)) = 0
y(x² - 8x + 4 + y²) = 0
y(x² - 2(4)(x) + (-4)² + 4 - (-4)² + y²) = 0
y( (x - 4)² + y² - 12) = 0
y = 0 or (x - 4)² + y² = 12
(Shown)
The other way is to directly equate the gradients
(y - 0)/(x - 1) = (-4y/(x² + y²) - 0)/( 4x/(x² + y²) - 1)
y/(x - 1) = -4y/(x² + y²) / (4x - x² - y²)/(x² + y²)
y/(x - 1) = -4y/(4x - x² - y²) = 4y/(x² + y² - 4x)
Solve accordingly to previous comment
y/(x - 1) = -4y/(x² + y²) / (4x - x² - y²)/(x² + y²)
y/(x - 1) = -4y/(4x - x² - y²) = 4y/(x² + y² - 4x)
Solve accordingly to previous comment
great thanks for all helps
Last part :
From earlier parts we can tell that the locus of P is a circle with centre (4,0i) and radius √12 units = 2√3 units
From earlier parts we can tell that the locus of P is a circle with centre (4,0i) and radius √12 units = 2√3 units
great thanks for all of your hard work,efforts and explanation
The locus of P and Q are points on the same circle , satisfying the equation, with PQ always passing through (1,0i)
So P and Q are always equidistant from the centre since you would form a radius if you draw either P or Q to the centre.
Now the radius from Q (the point 4/z) to the centre (4,0i) can be represented by |4 - 4/z|
i.e distance between two points or length of a line between 2 points.
Since the radius = 2√3 units,
|4 - 4/z| = 2√3
|4(1 - 1/z)| = 2√3
4|1 - 1/z| = 2√3
|1 - 1/z| = ½√3 (shown)
So P and Q are always equidistant from the centre since you would form a radius if you draw either P or Q to the centre.
Now the radius from Q (the point 4/z) to the centre (4,0i) can be represented by |4 - 4/z|
i.e distance between two points or length of a line between 2 points.
Since the radius = 2√3 units,
|4 - 4/z| = 2√3
|4(1 - 1/z)| = 2√3
4|1 - 1/z| = 2√3
|1 - 1/z| = ½√3 (shown)
appreciate for all your hard work