HELP: 1 Parameter Locus Question! (1 Viewer)

[kubekoo]

At a point P on the parabola, x^2=4ay (x^2 denoted x squared), a normal PK is drawn. From the vertex O a perpendicular OM is drawn to meet the normal at M. Show that the equation of the locus of M as P varies on the parabola is

x^4 - 2ayx^2 + (xy)^2 - ay^3 = 0

 

[Matthaeus]

equation of normal

x + py = 2ap + ap^3

now because its perpendicular to the normal and goes thru the origin

y - 0 = xp - 0 (same gradient as tangent)
y= xp

p= y/x

solving simultaneously

x + y^2/x = 2ay/x + ay^3/x^3

times both sides by x^3

x^4 + x^2y^2 = 2ayx^2 + ay^3

x^4 - 2ayx^2 + (xy)^2 - ay^3 = 0

as required

enjoy

 

[kubekoo]

ty~~ i never really though of that way~

 

[Sharky]

At a point P on the parabola, x^2=4ay (x^2 denoted x squared), a normal PK is drawn. From the vertex O a perpendicular OM is drawn to meet the normal at M. Show that the equation of the locus of M as P varies on the parabola is

x^4 - 2ayx^2 + (xy)^2 - ay^3 = 0

The Normal at P : x + py = 2ap + ap³

The perpendicular OM meeting PK at M : y = px

The point of intersection is thus:

Point M : x = ap(2+p²)/(1+p²) , y = ap²(2+p²)/(1+p²)

Now, the only way that i can show that x^4 - 2ax²y + x²y² - ay³ = 0 is the cartesian equation for M is to substitute these two lovely values for x and y into that equation. I just so happened to be bored enough to do that, and it does in fact turn out to equal zero.

Hence we have shown that the locus of M is in fact the given equation.

 

[Sharky]

Nice work ChiQui, wish i had've seen that. would've saved me all my toil and grief.
But hey, i am renound for the abstract method of obtaining the correct solution.

 

[Matthaeus]

Hehe glad i could help, it must've been one big nightmare subbing it in and evaluating, its quite a feat doing that man im scared thinking about all the ps everywhere.