Help with parametric question! (1 Viewer)

[Munkiie]

I need help with this question. I don't know what to do with the angle??

You are given the tangents at P and Q (which I found was y=px - ap^2 & y=qx -aq^2) intersect at an angle of 45. Show that p-q= 1 + pq.

Thanks in advance.

 

[asianese]

This question is talking about the angle between two lines.

 

[Munkiie]

Omg I forgot all about that formula. Thank you very much.

 

[asianese]

 

[Munkiie]

Ok I'm stuck again... I'm terrible at this =.=

It's the next part of that question. It says:

By evaluating the espression x^2 = 4ay at T (which is pt of intersection I found is (a(p+q),apq)), or otherwise, find the locus of T when the tangents at P and Q intersect.

I tried using x=a(p+q), y=apq as parametric equations but I'm not getting the right answer
Please help

 

[Sy123]

Ok I'm stuck again... I'm terrible at this =.=

It's the next part of that question. It says:

By evaluating the espression x^2 = 4ay at T (which is pt of intersection I found is (a(p+q),apq)), or otherwise, find the locus of T when the tangents at P and Q intersect.

I tried using x=a(p+q), y=apq as parametric equations but I'm not getting the right answer
Please help

A locus cannot be defined unless we have a relationship between p and q (in this case)

There must be a previous part of the question, i.e. assume pq=2 or perhaps they are tangents from a focal chord?
There is no way at all to achieve an answer (in only x and y) without some relationship given about p and q

 

[Munkiie]

Well the only relation I got was p-q=1 +pq
But I still don't know know what to do with it...

 

[rolpsy]

I need help with this question. I don't know what to do with the angle??

You are given the tangents at P and Q (which I found was y=px - ap^2 & y=qx -aq^2) intersect at an angle of 45. Show that p-q= 1 + pq.

Thanks in advance.

There should really be absolute value signs around that:   (unless some other condition is placed on p and q), because the angle between two lines takes the absolute value:



Note how now the expression is symmetric in p and q, i.e., they can be exchanged without changing the relation. (|p – q| = |q – p|)


Now, we need to eliminate p and q from



using

absolute value signs are generally an indication to square, after which we can use    to insert x

 

[Munkiie]

There should really be absolute value signs around that: * (unless some other condition is placed on p and q), because the angle between two lines takes the absolute value:



Note how now the expression is symmetric in p and q, i.e., they can be exchanged without changing the relation. (|p – q| = |q – p|)


Now, we need to eliminate p and q from



using

absolute value signs are generally an indication to square, after which we can use ** to insert x

Thanks it's really helped