Locus question (1 Viewer)

[Jaymee]

I was doin sum revision for 2u nd got stuck on this q. Ny help with it would b appreciated. if possible culd u explain wat ur doin as u go.

a) find the equation of the parabola with focus (-1, 6) and vertex (-1, 4)
b) take the origin and the point A(2a,0) as two fixed points. Assume P(x,y) moves so that PA perpendicular to PO. Show that the locus of P is a circle on OA as diameter. (this is the part i don't get)

 

[Loz_metalhead]

do you have the answers? It will help.

 

[Jaymee]

nope

 

[rama_v]

I was doin sum revision for 2u nd got stuck on this q. Ny help with it would b appreciated. if possible culd u explain wat ur doin as u go.

a) find the equation of the parabola with focus (-1, 6) and vertex (-1, 4)
b) take the origin and the point A(2a,0) as two fixed points. Assume P(x,y) moves so that PA perpendicular to PO. Show that the locus of P is a circle on OA as diameter. (this is the part i don't get)

A diagram really helps. Becuase in this case the parabola is concave up, the equation is of the form (x-h)²= 4a(y-k) where (h, k) is the vertex. so it becomes (x+1)² = 4a(y-4)

Since the focal length is 2, a = 2, so:
(x+1)² = 8(y-4) is the required equation of the parabola.

 

[vafa]

answer to the first question:
Focus(-1,6) and Vertex(-1,4)
one of the general equations of a parbola is in the form:
(x-h)^2=4a(y-k)
where:
vertex(h,k)
focus(h,k+a)
y=k-a the equation of the directrix
x=h the equation of the axis of symetry of the parabola

by comparing this general equation and given information, you will notice that:
h=-1
k=4
k+a=6
hence a=2
then you just put the values of h, k and a in the general equation of the parabola:
(x+1)^2=8(y-4)
this answer is right but if you like to expand this you can :
x^2+2x+1-8y+32=0
hence you will get:
x^2+2x-8y+33=0


answer to the second question:
O(0,0), A(2a,o) , P(x,y)

because the question says PA and Po are prependicular then the multiplication of gradient of PA and Po=-1
gradient of PA=y/(x-2a)
gradient of PO=y/x

hence:
y^2/(x^2-2ax)=-1
y^2=-x^2+2ax
x^2-2ax+y^2=0
(x-a)^2+(y-0)^2=a^2
center(a,0) and radius=a
diameter=twice of radius
diameter=2a

length os Ao=2a

hence the length of OA is the same as the length of the diameter then we say that oA is the diameter of the circle.

 

[Jaymee]

answer to the first question:
Focus(-1,6) and Vertex(-1,4)
one of the general equations of a parbola is in the form:
(x-h)^2=4a(y-k)
where:
vertex(h,k)
focus(h,k+a)
y=k-a the equation of the directrix
x=h the equation of the axis of symetry of the parabola

by comparing this general equation and given information, you will notice that:
h=-1
k=4
k+a=6
hence a=2
then you just put the values of h, k and a in the general equation of the parabola:
(x+1)^2=8(y-4)
this answer is right but if you like to expand this you can :
x^2+2x+1-8y+32=0
hence you will get:
x^2+2x-8y+33=0


answer to the second question:
O(0,0), A(2a,o) , P(x,y)

because the question says PA and Po are prependicular then the multiplication of gradient of PA and Po=-1
gradient of PA=y/(x-2a)
gradient of PO=y/x

hence:
y^2/(x^2-2ax)=-1
y^2=-x^2+2ax
x^2-2ax+y^2=0
(x-a)^2+(y-0)^2=a^2
center(a,0) and radius=a
diameter=twice of radius
diameter=2a

length os Ao=2a

hence the length of OA is the same as the length of the diameter then we say that oA is the diameter of the circle.

how did u get the gradient of PA?
i havent done this stuff in ages, so i'm a bit slow.

 

[vafa]

if A(a,b) and B(c,d)
the gradientd of line Ab=(d-b)/(c-a) or
the gradient of line Ab=(b-d)/(a-c)

 

[Jaymee]

okay, i get ya. just do the whole rise/run thing. thanks.

 

[vafa]

i know that A(2a,0) and p(x,y) from the given information:
Gradient of PA=(y-0)/(x-2a) or
gradient oF Pa=(0-y)/(2a-x)

there is no difference between them. in order to understand this, you need to know the gradient formula as i mentioned in my previous post; you can have a look at your book and understand some of the basic concepts which you need to understand when you need to solve locus quastions such as: gradient formula, prependicular distance form a line formula and etc.

 

[rocafella32]

this is off the sub, but yeah i was also doin some maths studying for half yrlys and i keep thinking to myself there must be an easier way to study maths than Q's after Q's after Q's......is there??

 

[SoulSearcher]

Not really.

 

[vafa]

i am not someone to advise you to how study mathematics but i can share what i do in order to do well in my exams.

i never memorise formulas or solutions, i just understand them.

I read through the text book, master the concept then i do some questions which are in three different levels:foundation, development and extension

in my perspective Cambridge books are much better than Mathematics in focus or fitzapetrik because they have questions in 3 different levels.

 

[SoulSearcher]

IMO, Maths In Focus questions are just way too easy, good to learn a topic, but not good for challenging the mind of the student in terms of question difficulty. Fitzpatrick and Cambridge are both good for 2 unit maths.

 

[rocafella32]

What about excel and and those excel revise in a month book thingys

 

[vafa]

when you study cambridge or other similar text books , you will understand all concepts and finally you are sure 100 percent about them.

but i personally think that excell books are very complicated if u want to do a work for the first time but obviously they can be useful to revise what you have learnt.

 

[Riviet]

I personally think the excel study guides for mathematics have well explained examples, but lack the actual excercises that you can get from textbooks such as fitzpatrick and cambridge. The excel series is generally good for revising over topics, but not for practising questions and problems.

 

[Mumma]

Umm, for the second part cant you just say that P must lie on a circle with OA as diameter due to the fact that the line OA is a chord subtending a right angle at the circumference?

 

[Riviet]

Umm, for the second part cant you just say that P must lie on a circle with OA as diameter due to the fact that the line OA is a chord subtending a right angle at the circumference?

It's tempting to say that, since it's a circle geometry theorem, but we are asked to show it, therefore we can't just state a theorem, unless the question asked us to explain why the locus is that.