SOLUTION: P (2p,p^2) is a variable point on the parabola x^2 = 4y. N is the foot of the perpendicular form P to the x axis NR is perpendicular to OP. a. find the equation of OP b. find t

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: P (2p,p^2) is a variable point on the parabola x^2 = 4y. N is the foot of the perpendicular form P to the x axis NR is perpendicular to OP. a. find the equation of OP b. find t      Log On


   

Question 1039422: P (2p,p^2) is a variable point on the parabola x^2 = 4y. N is the foot of the perpendicular form P to the x axis NR is perpendicular to OP.
a. find the equation of OP
b. find the equation of NR
c. show that the R has coordinates (8p/p^2 +4 , 4p^2/p^2+4)
d. show that the locus R is a circle and state its centre and radius
NEED HELP WITH PART C AND D!!!
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
a. {{OP}}} connect O%280%2C0%29 and P%282p%2Cp%5E2%29 ,
so its slope is p%5E2%2F2p=p%2F2 and its y-intercept is 0 .
The equation of OP is y=%28p%2F2%29x

b. The foot of the perpendicular form P%282p%2Cp%5E2%29 to the x-axis is
N%282p%2C0%29 , and the slope of a line perpendicular to OP is
-1%2F%28p%2F2%29=-2%2Fp .
So, the equation of NR in point-slope form, based on point N%282p%2C0%29 is
y=%28-2%2Fp%29%28x-2p%29

c. you did not tell me where point R is, but from its coordinates, I see that is a point on line OP .
so, R must be the intersecion of NR and OP.
Hence, its coordinates are the solution to
system%28y=%28p%2F2%29x%2Cy=%28-2%2Fp%29%28x-2p%29%29 .
Using substitution, we start trying to find x from %28-2%2Fp%29%28x-2p%29=%28p%2F2%29x .
Multiplying both sides times -2p we get
%28-2p%29%28-2%2Fp%29%28x-2p%29=%28-2p%29%28p%2F2%29x .
4%28x-2p%29=-p%5E2x
4x-8p=-p%5E2x
4x%2Bp%5E2x=8p
%284%2Bp%5E2%29x=8p
highlight%28x=8p%2F%284%2Bp%5E2%29%29 .
Then, substituting the value found for x into y=%28p%2F2%29x we get
y=%28p%2F2%29%288p%2F%284%2Bp%5E2%29%29
highlight%28y=4p%5E2%2F%284%2Bp%5E2%29%29

d. R%288p%2F%284%2Bp%5E2%29%2C4p%5E2%2F%284%2Bp%5E2%29%29 is symmetrical with respect to the y-axis,
because substituting -p for p you get point S%28-8p%2F%284%2Bp%5E2%29%2C4p%5E2%2F%284%2Bp%5E2%29%29 ,
the reflection of R%288p%2F%284%2Bp%5E2%29%2C4p%5E2%2F%284%2Bp%5E2%29%29 across the y-axis.
Making p=0 gives us the point O%280%2C0%29 , which is part of the locus or R .
All the points R%288p%2F%284%2Bp%5E2%29%2C4p%5E2%2F%284%2Bp%5E2%29%29 have a non-negative y-coordinate,
y=4p%5E2%2F%284%2Bp%5E2%29%3E=0 .
So the circle should be centered on the y-axis,
pass though O%280%2C0%29 ,
and otherwise be above the y-axis.
Such a circle has a radius r, and is centered at C%280%2Cr%29 .
All we need to do is find r .
The equation of such a circle would be
%28x-0%29%5E2%2B%28y-r%29%5E2=r%5E2 <---> x%5E2%2By%5E2-2ry%2Br%5E2=r%5E2 <---> x%5E2%2By%5E2-2ry=0 <---> x%5E2%2By%5E2=2ry .
Substituting x=8p%2F%284%2Bp%5E2%29 and y=4p%5E2%2F%284%2Bp%5E2%29 , we can find r .

64p%5E2%2F%284%2Bp%5E2%29%5E2%2B16p%5E4%2F%284%2Bp%5E2%29%5E2+=+8rp%5E2%2F%284%2Bp%5E2%29
64p%5E2%2B16p%5E4+=+8rp%5E2%284%2Bp%5E2%29
16p%5E2%284%2Bp%5E2%29+=+8rp%5E2%284%2Bp%5E2%29
16p%5E2=+8rp%5E2
16p%5E2%2F8p%5E2=r ---> highlight%28r=2%29 .
The locus of R is a circle of radius 2 centered at C%280%2C2%29 .

NOTES:
The equations to graph that circle: system%28x=8p%2F%284%2Bp%5E2%29%2Cy=4p%5E2%2F%284%2Bp%5E2%29%29
can be called parametric equations.
Parametric equations allow us to show a relation that is not a function,
so we can express a relation that graphs as a curve we call a parametric curve.
They are very useful if would be difficult to express with just x and y .
A third variable called parameter (in this case p ) is uaed,
and we define x and y as functionc of that parameter.
Many times, we try to "eliminate the parameter",
to get to a relation between just x and y ,
be able to understand the relation between x and y more easily.
One way to do that is to solve one equation for the parameter,
getting an expression for the parameter in terms of x and y that can be substituted for the parameter in the other equation for the other variable.
That would have been complicated in this case, but fortunately we were told it was a circle and all we had to do is prove it and find center and radius.
That made it easier.
Maybe the way I went about it is not the expected way, but it was the easiest way I could figure out.
If you find out about some easier way, let me know.