Loci problem

[Sonal7]

I am having trouble with part d of the question. I have not learnt anything about ellipses, and i am wondering if they might be referring to
an ellipse in the answer. I just understand why they sub in x and -2yi, i know that is the midpoint of the circle is 4, and -2, might this be the reason.
It is given that the line arg(w-6) =alpha passes through the centre, but one cant assume the line is passing through the origin.
It turns out it is. I am confused as i thought that arg(w-6) is a line that goes from Re 6 to 4,-2 and then passes through the circle to q whose
co-ordinates we must fine.

 

[Deleted member 4993]

Please share your work regarding parts (a) through (c)

 

[Dr.Peterson]

I tend to agree with you; but I can't tell what they are doing without seeing the solution to 11c, where they must have defined m, n, and b.

 

[Sonal7]

a) [MATH]\mid z-6i\mid=2\mid z -3\mid[/MATH]b)
[MATH]\mid x+yi-6i\mid=2\mid x + yi-3\mid[/MATH][MATH]x^2+(y-6)^6=4(x-3)^2+4y^2[/MATH][MATH]=y^2+x^2+8x+4y=0[/MATH]c)
The above equation can be written as:
[MATH] (x-4)^2+(y+2)^2= (2\sqrt[2]{5})[/MATH]Therefore the centre of the circle is 4,-2

[MATH]Tan^-1(\frac{-1}{2})[/MATH]=5.82 (2 d.p).

 

[Sonal7]

I cant see that they have defined m, n and b, I think they are alluding to a linear equation perhaps. I have no idea. I cant see how arg (w-6) passes through the origin, and its been shown that it does. At Q I would think that the the values of x and -2y could be sub in to the equation of the circle but I dont know how they know what that is. I have drawn a diagram.

 

[Dr.Peterson]

I got a different value for alpha; I could be wrong, but it doesn't look like they have justified their answer.

Check their answer by seeing whether it satisfies all the conditions. I think they are wrong, but I can't spend more time just now.

 

[Sonal7]

Yes i did but as its in the 4th quadrant you have to work it out so that its 2 pi-theta. I am not sure why they have done this, as in Argrand diagram the convention is the solution is -theta if the point is in the fourth quadrant. One is measuring theta compared to the horizontal. I just realised that as its a loci problem one does need to know which angle is under Q. That might be a big clue. I need to go away and think.

 

[Sonal7]

I think they are wrong too. I wont spend more time on this as this problem is likely to unravel in sleep. I am so tired bringing questions to the forum for which the answers to the questions turn out to be wrong. Must trust my instinct.

 

[Deleted member 4993]

a) [MATH]\mid z-6i\mid=2\mid z -3\mid[/MATH]b)
[MATH]\mid x+yi-6i\mid=2\mid x + yi-3\mid[/MATH][MATH]x^2+(y-6)^6=4(x-3)^2+4y^2[/MATH][MATH]=y^2+x^2+8x+4y=0[/MATH]c)
The above equation can be written as:
[MATH] (x-4)^2+(y+2)^2= (2\sqrt[2]{5})[/MATH]Therefore the centre of the circle is 4,-2

[MATH]Tan^-1(\frac{-1}{2})[/MATH]=5.82 (2 d.p).

Is this the "work" shown in the book?

It has lots of "mistakes"! Go over the calculations carefully!

 

[Sonal7]

Is this the "work" shown in the book?

It has lots of "mistakes"! Go over the calculations carefully!

No its correct except for a sign. It should be -8x. I have just not shown all the steps but it works out okay as the questions asks 'show'. I worked out the equation as per the questions which is correct.

 

[Dr.Peterson]

Here's the check: If Q = (8, -4), i.e. w = 8 - 4i, then arg(w-6) = arg(8 - 4i - 6) = arg(2 - 4i) = arctan(-2), not arctan(-1/2). So it doesn't satisfy the equation of the line with alpha being what they claim it is.

 

[Deleted member 4993]

No its correct except for a sign. It should be -8x. I have just not shown all the steps but it works out okay as the questions asks 'show'. I worked out the equation as per the questions which is correct.

Really?

In your work you show a term (y - 6)⁶

Where did that come from?

And some more......

 

[Sonal7]

Really?

In your work you show a term (y - 6)⁶

Where did that come from?

And some more......

Its is a typo. I checked these and it works out ok.

Here are the steps:
[MATH] \mid {x+yi-6}\mid=2 \mid{x+yi-3}\mid [/MATH][MATH]x^2 +(y-6)^2=4(x-3)^2+4y^2[/MATH][MATH]x^2+y^2-12y+36=4(x^2-6x+9)+4y^2[/MATH][MATH]x^2+y^2-12y+36=4x^2-24x+36+4y^2[/MATH][MATH]3y^2+3x^2-24x+12y=0[/MATH][MATH]y^2+x^2-8x+4y=0[/MATH]This is the equation required, so shown.