Locus Questions (1 Viewer)

[Seraph]

Hey ppl , with locus is there an easier way to do the quadratic type problems

e.g Find the equation of the locus of a point that moves so its equidistant from the point (0,-3) and the line y = 3

I know you can use the distance formula , but someone told me you can use a parabola equation... how do i do this?

 

[crazylilmonkee]

i only ever got taught the distance formula..
its not that hard

 

[Seraph]

actually i did use the distance formula For that Particular example i gave above
Find the equation of the locus of a point that moves so its equidistant from the point (0,-3) and the line y = 3


i ended up gettin just 12y

answer says x^2 + 12y
im not sure what i did wrong

after gettin rid of Square roots
(x-0)^2 + (y+3)^2 = (x-0)^2 + (y-3)^2
x^2+y^2+6y+9 = x^2 + y^2 - 6y + 9

 

[Winston]

u can use the parabola, actually to solve that easily like in some circumstances


you see the point is


(0,-3) and y = 3 is the directrix


we always know that

from the focus point to the vertex and from the vertex to the directrix is equal in distance, that's always

so in this scenario if we plot it on a cartesian plane we know that this is going to be a general parabola at vertex of (0,0)

so therefore it's in the form of


x^2 = 4ay


therefore you know the characterics of this actual general parabola which is


focus (0,a)
directrix y = -a
vertex (0,0)

thefore we know that a = -3


so if you plug that into the actual general equation bit you'll have 4(-3) = -12

thus it's equation is

x^2 = -12y

and the minus sign indicates it's a upside down parabola.

 

[Lexicographer]

Very nice. But you mean concave down, there's no such thing as an "upside down" parabola.

(first post in maths since exam)

 

[psycho_mushy]

DUDE! MATHS IS OVER! GET OVER IT!

 

[Winston]

Originally posted by Lexicographer
Very nice. But you mean concave down, there's no such thing as an "upside down" parabola.

(first post in maths since exam)

psssh look who's post stalking now, such a kid i swear.

 

[Seraph]

Okay i have 2 more questions if anyones there

consider 15-8x-x^2 in form a - (x+b)^2

Once u put that in that form its 31 - (x+4)^2

but then it says find the maximum value of 15-8x-x^2 and the value for x for which it occurs.. i dont understand this..


also the following equation 8y = x^2 - 6x - 3
In form (x-h)^2 = 4a(y-k) its (x+3)^2 = 2(y-1.5)
V(-3,1.5)
focus (-3,0.5) rite? is The 2 In the Equation the focal LENGTH or the focus Point at y ?

and heres what i dont get the directrix is apparently y = -3.5 but how does that work? if the distance is 0.5 then from the Vertex shouldnt the directrix be y = 1 ???

 

[Winston]

Originally posted by Seraph
Okay i have 2 more questions if anyones there

consider 15-8x-x^2 in form a - (x+b)^2

Once u put that in that form its 31 - (x+4)^2

but then it says find the maximum value of 15-8x-x^2 and the value for x for which it occurs.. i dont understand this..


also the following equation 8y = x^2 - 6x - 3
In form (x-h)^2 = 4a(y-k) its (x+3)^2 = 2(y-1.5)
V(-3,1.5)
focus (-3,0.5) rite? is The 2 In the Equation the focal LENGTH or the focus Point at y ?

and heres what i dont get the directrix is apparently y = -3.5 but how does that work? if the distance is 0.5 then from the Vertex shouldnt the directrix be y = 1 ???

First ques, maximum value is just another way of saying the maximum 'y' value, if u draw up a curve yea ur just finding the value at which its the highest at, so in this case when x = 0 y = 15, so the highest value is 15 when x = 0


hmmm damn your making me remmeber maths

hmmm ur equation looks alrite

but it should be (x-3)^2 that's the error that stuffed u up

 

[Seraph]

hmm thats odd it says the Maximum value is 31 when x = - 4

and how did u draw up that curve if u cant factorize it

oh while im here might as well ask this to say with an equation
x^2 = 8y
Why is the axis of symmetry x = 0? why is it not y= 0

and how do i find axis of symmetry for (x-1)^2 = 4(y+2)

 

[Winston]

Originally posted by Seraph
hmm thats odd it says the Maximum value is 31 when x = - 4

and how did u draw up that curve if u cant factorize it

oh while im here might as well ask this to say with an equation
x^2 = 8y
Why is the axis of symmetry x = 0? why is it not y= 0

and how do i find axis of symmetry for (x-1)^2 = 4(y+2)

Oh yes it is correct i didnt look at the factorised form ahuh becoz wen u substitute in -4 ull get y = 31, thats wats ur aim is, to find the largest y value


axis of symmetry hmmm im not sure if i remember, isnt it -b/2a? :S