huh???

[egg]

please help me with Q1e)
I got a to d) just not e)

thanks

[keltingmeith]

I must say, a very interesting question. I wouldn't expect it on an exam, but it's pretty good for interest! Also, in future cases like this, I would definitely suggest giving what you got for the previous parts of the question - they definitely help.

So, for this question, I used my solution that the point for M is given by

Now, let's first consider the case for when

If we take the y-value of M and sub in this value for b, we can see that , which means that when , all of the y-values for the co-ordinate M must be 0. This gives us that for , . We can also note that the range of the x value is . This means that this set of loci are all those x values. Using set notation, for , . In cartesian form,

What about for when ? Well, using what we did before, we saw that for , , and similarly, for , . So, for , we will get an elipse, given by the set of points , where

[egg]

oh ok thank you

I wasn't expecting it to be this difficult, but how did you get:

We can also note that the range of the x value is . This means that this set of loci are all those x values. Using set notation, for , . In cartesian form,

What about for when ? Well, using what we did before, we saw that for , , and similarly, for , . So, for , we will get an elipse, given by the set of points , where

[keltingmeith]

So, we know that the range of x values is given by , because this is what is.

To find the range of values that this can take, just do what you would for anything - graph against , and see what values come up for . You should see that can take on any value between and .

This is how we got the bit in set notation. For the cartesian form, we know that (as proven earlier), so that means that we have the equation , and the domain of this equation is given by the range of x values, since this cartesian equation is built off of the parametric equations and .

For the second part, if I were to visualise the tracing of the loci made by M, we'd get another ellipse OR a circle. To tell the difference, we check whether or not . In this case, or for the locus of M to give us a circle. (this is required for the next part of the question, I noticed) However, b was already defined such that , so b cannot take those values, and we must get an ellipse.

[egg]

oh yea i see but the ans says its 0<x<1/2???
and cause with the b not equal to 1/sqrt2 how did they get the formula: 4x^2 + (4b^2y^2)/(2b^2-1)^2 = 1? yup and that shows that it is an eclipse which is what you were saying before...

[keltingmeith]

Whoops - I missed the restrictions on theta! Hahah, this is why you read the whole question and highlight important bits.

If you graph x against theta, as I said before, and consider the restrictions to the domain of theta, you should get the range 0 < x < 1/2. As for their formula, what they've done is used the pythagorean identity.

So, if you say that:


And remember that:



You can rearrange to get and and put them into the pythagorean identity to get their equation.

[egg]

OMG it makes so much sense now thank thank you thank you

just for the last part of the question:
it says to find the value(s) for b which the locus is part of the circle:

so what i thought was i would have to use that eclipse formula from previous part: 4x^2 + (4b^2y^2)/(2b^2-1)^2 = 1
and for it to be a circle a = b and therefore i made:
(2b^2-1)^2 = 1 cause a is 1 right?
2b^2-1 = 1
i got b = plus minus 1 if i did this, but the ans says 1/2???

[keltingmeith]

Very close! Remember the coefficients in front of and . Considering those, you'll actually find the values of a and b are:




(note: I've used and to differentiate between the actual variable b)

Everything you else you did was right - just remember that the x and y terms must be by themselves if you consider this case, because in the formula we're using:



x and y are by themselves, with a and b as the denominators.

EDIT: They might be suspicious, but I'm just a normal citizen providing good-will to struggling year 12s, so I think this is best left to the moderators and I'ma just keep on answering questions.