General Parametric Equations (1 Viewer)

[Managore]

Okay I hate this chapter, to me it makes no sense whatsoever. I can create the cartesian equation from two x= and y= equations of course, but I can not understand the tangents to points P, Q, and anything like that. Could someone give me some general tips or a brief explanation?

 

[jumb]

Know how to differentiate.....

I'm done.

 

[Managore]

Umm.. that'll help me find their gradients.. then would I use y-y1=m(x-x1) or are there other tips/tricks/etc I can do?

 

[smallcattle]

tangent to P and Q???

just find the gradient of the curve, sub in the x in P or Q, then use point gradient formula to get the equation of the tangent of the curve at P or Q???

 

[BillyMak]

Do you know how to find locuses?

EDIT: loci, even?

 

[Managore]

Yes, of course I know how to find locuses.. I just never understood how using a third dynamic term worked.

 

[smallcattle]

Do you know how to find locuses?

EDIT: loci, even?

how to find locus and loci??

 

[Managore]

uhh it depends on the given information. eg they give you directrix and focus, and you find a and the vertex and sub into

(x-h)^2=4a(y-k) where the vertex is (h, k)

 

[BillyMak]

They're the words I am looking for

Ok, so if I were to say to you, P(2ap,ap²) and Q(2aq,aq²) are two points on the parabola x² = 4ay, find the locus of the path traced by the intersecting normals corresponding to each point if the normal at P goes through the point S(0,a), could you find the locus?

 

[jumb]

find the equations of each normal, then find a by itelf on each normal, and solve them simultaneously.

I think.

 

[BillyMak]

You would find the point of intersection of the two normals, using simultaneous equations. The points would include ps and qs in them, and you don't want that if you are trying to get the locus.
You would then substitute the point (0,a) into the equation of the normal through P, and get an expression for p. Sub that into your intersection point, and you now have the intersection point in terms of q and a, which is good since a is a constant.

You then say x = the x coordinate (in terms of q and a) and y = the y coordinate (in terms of q and a) then solve them simultaneously, eliminating q, which gives you the cartesian equation of the locus.

 

[BillyMak]

P.S. Don't try to do this question, it was just something I made up on the spot and since the normal through P is a fixed line the locus of the intersection is obviously the the fixed line. As such, it shouldn't be taken seriously

 

[Managore]

oh.. dear.. god.. I am going to fail. Is that the only question type they'll give us? or are there a variety of nonsensical questions like that one?

 

[BillyMak]

They won't give you a question in a big block like that, chances are they'll put it in parts, like (i) could be find the equation of the normals, (ii) could be find their intersection point, (iii) could be given a bit of info, find the locus of the intersection.

P.S, a reminder that it is a dodgy question so don't attempt to do it (although the locus is x = 0)

 

[smallcattle]

what is locus???? isnt locus the point where the vertex of the parabola +- the focal length??

 

[Managore]

thats focus.. locus is the parapola.. a locus is something that satisfies given criteria (eg equidistant from a line and focus)

 

[smallcattle]

oh oh..........OMG i cant even distinguish between the two..

 

[BillyMak]

Yeah, you might want to clarify the difference before tomorrow