2006 HSC Question 2(d)ii) (1 Viewer)

[lukezy]

After going through this past paper, and comparing my answer with the worked solution i still don't understand where they got their answer from. :S

The equation |z-1-3i| + |z-9-3i| = 10 corresponds to an ellipse in the Argand diagram.
After evaluating the centre sketch the ellipse, stating the lengths of the major and minor axes

The work solution claims the equation of the ellipse to be:
[(x-5)^2] / 25 + [(y-3)^2] / 9 = 1
and yes from this we can draw the diagram...
but they show no working out and i have no idea as to where they got this from.

Help would be much appreciated
Thankyou

 

[Poad]

It's of the form PS + PS' = 2a (i.e another definition of an ellipse)

From the equation 2a = 10, .:. a = 5
And the centre is the midpoint of (1,3) and (9,3) -> (5,3)
And, uh, I can't quite remember how to find the other things (b and e).

 

[vds700]

to find b, you sub x = 5 into

|z-1-3i| + |z-9-3i| = 10 as this is the highest point of the ellipse
l(x-1) +i(y-3)l + l(x-9) +i(y-3)l = 10 let x = 5
l4 + i(y-3)l + l(-4 + i(y-3)l =10
sqrt(16 + (y-3)^2) + sqrt(16 + (y-3)^2) =10
2sqrt(16 + (y-3)^2) = 10
4(16 + (y-3)^2) = 100
16 + (y-3)^2 = 25
(y-3)^2 = 9
y-3 = +-3
therefore y = 0, 6

therefore minor axis length = 6, b = 3

 

[tommykins]

回复: 2006 HSC Question 2(d)ii)

After going through this past paper, and comparing my answer with the worked solution i still don't understand where they got their answer from. :S

The equation |z-1-3i| + |z-9-3i| = 10 corresponds to an ellipse in the Argand diagram.
After evaluating the centre sketch the ellipse, stating the lengths of the major and minor axes

The work solution claims the equation of the ellipse to be:
[(x-5)^2] / 25 + [(y-3)^2] / 9 = 1
and yes from this we can draw the diagram...
but they show no working out and i have no idea as to where they got this from.

Help would be much appreciated
Thankyou

Definition of ellipse shows that the two points are the foci of the ellipse. The midpoint of these two points are therefore the centre z = 5+3i)

The major axis can be found by simply realising that at the y-axis, the horizontal distance from P1 to P2 is 10 units and therefore, the length of the major axis is 10.

Minor axis can be found using pythagora's where the hypotenuse = 5 and the distance from centre = 4.

The argument question is simple, drawing the ellipse you only get 0 < arg(z) < pi/2 (theres equal signs).

I loved this question, was in my half yearlies, brilliant.

 

[lukezy]

Thank you to all three of you
All helped and i understand ^^

well.... ill get back to my papers now
Take care XD