AS 91577 · 2023 paper · dividing in polar form, a modulus in k, an argument, cube roots, and a tangent to a circle
Question 2(e) Merit to Excellence
Practises Loci of complex numbers
Five separate parts: a quotient of two numbers in polar form, the modulus of a difference, an argument found from an equation, the cube roots of a negative number, and a line that touches a circular locus.
The straight line with equation y = mx − 1, where m is a real constant and m > 0, is a tangent to the locus described by |z − 2 + i| = √3.
Find the Cartesian equation of the locus AND the value of m.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 Put z = x + yi and write out the modulus.
|(x − 2) + (y + 1)i| = √3, so √((x − 2)2 + (y + 1)2) = √3 -
2 Square both sides. What shape is it?
(x − 2)2 + (y + 1)2 = 3: a circle, centre 2 − i, radius √3. -
3 Where does the line meet it? Substitute y = mx − 1.
(x − 2)2 + (mx − 1 + 1)2 = 3
x2 − 4x + 4 + m2x2 = 3
(1 + m2)x2 − 4x + 1 = 0 -
4 A tangent touches at exactly one point. What does that say about this quadratic?
It has one repeated root, so its discriminant is zero:
16 − 4(1 + m2)(1) = 0, so 1 + m2 = 4 and m2 = 3. -
5 Which sign?
The question says m > 0, so m = √3. (It touches at x = ½.)
What the marker wanted
- Achieved The modulus written out after z = x + yi (step 1).
- Merit The line substituted in to give (1 + m2)x2 − 4x + 1 = 0.
- Excellence m, with a valid, clear reason (a zero discriminant): E8. A full answer with one minor error: E7. Giving m = ±√3 is not penalised.
In our words, from NZQA’s 2023 marking schedule.
How did it go?
Got it moves it further out. Review brings it back tomorrow.