Parabola help!! 2 questions! (1 Viewer)

[studentcheese]

OMG this is so embarrassing. Anyways, I seem to be struggling with two parabola questions:

1) the tangent at P(2ap,ap^2) on the parabola x^2=4ay cuts the y axis at T. M is a point on PT that divides PT internally in the ratio 2:1. Show that the cartesian equation of the locus of M is a parabola whose vertex is the origin. What is the focal length of the parabola?

gradient at P... y = x^2/4a = dy/dx = x/2a
gradient of P = p

equation of p.....

p = (y-ap^2)/(x-2ap)
px - 2ap^2 = y - ap^2
px - ap^2 = y

point T.....
when x = 0, y = -ap^2
T = (0, -ap^2)

okay so how do I do I find out the coordinates of M? :mad1:

2) A and B are two points on the parabola x^2 = 4ay, with parameters p, p - 1, respectively. At and Bt are tangents. Show that the locus of T is x^2 = 4a(y+a/4).

I have no idea what this question is trying to get.

Thanks in advance

 

[shady145]

use the formula from the prelim course
dividing lines internally
M((mx2+nx1/m+n),(my2+ny1/m+n))

 

[Michaelmoo]

Ok your on the right track.

1) the equation for a division of a line interval is given by x = (mx2 + nx1)/(m+n) and y = (my2 + ny1)/(m + n).

I trust you've seen these equations before? Bsically you divide a line interval from (x1, y1) to (x2, y2) in the ratio of m:n internally by plugging figures into that. If you want to divide externally, you let m = -m but we don't need to consider this for now.

Now, it is important which point you consider to be x1 and x2. Basically it's in the order they are read aloud. For example, in this case its the line interval PT, therfore the P will represent the 1 values and T will represent the 2 values. So:

(x1, y1) = P = (2ap,ap^2)

(x2, y2) = T = (0, -ap^2)

m:n = 2:1

Now we simply do a substitution:

x = (mx2 + nx1)/(m+n)

x = (2(0) + 1(2ap))/(2+1)

x = 2ap/3 ................(1)


similarly:

y = (my2 + ny1)/(m + n)

y = (2(-ap^2) + 1(ap^2))/(2 + 1).

y = (-ap^2)/3 ........ (2)


Ok. so We've found our 2 points, now let us find the locus, by solving simultaneously to eliminate the parameter p

Now from (1) --> p = 3x/2a. Square both sides: p^2 = 9(x^2)/(4a^2) ...... (3)

Notice how we square both sides to try and get a p^2, to eliminate the p into equation (2).

Now, sub (3) into (2)

y = (-a(9x^2)/4a^2/3

y = -9x^2/12a

9x^2 = -12ay

x^2 = (4/3)ay

Which is a parabola with vertex (0,0) and focal length a/3

2)Ok so you know that the parabola x^2 = 4ay, has points (2aq, aq^2) where q is a varying parameter. Now:

A (2ap, ap^2)

B [2a(p-1), a(p-1)^2]

Ok, with the information, we're given, we know we that the tangetnts at A and B intersect at the point t. So let us find the equation of the tangents at A and B and solve for t.

You know how to do this so I'm just going to skip to the equation:

y = px - ap^2 .......(1)

y = (p-1)x - a(p-1)^2 .......... (2)

Noq (1) - (2) will eliminate y to give us an x-coordinate for T

0 = px - (p-1)x - ap^2 + a(p-1)^2

0 = x - ap^2 + a(p^2 - 2p + 1)

0 = x - ap^2 + ap^2 -2ap + a

x = 2ap - a -----------------(a)

Now lets find the y- coordinate of t, by subbing in the x-coordinate into either equation (1) or (2)

y = px - ap^2

y = p(2ap-a) - ap^2

y = 2ap^2 -ap -ap^2

y = ap^2 - ap = ap (p-1) ------------------(b)

now to solve simultaneously, look for methods to eliminate p. from (a) (the question also gives you a clue)

x = 2ap - a

x = a (2p-1)

x/a = (2p - 1)

x^2 / a^2 = (2p - 1)^2 = 4p^2 -4p + 1

x^2/ a = 4ap^2 -4ap + 1..................(c)

now y = ap^2 - ap

4y = 4ap^2 - 4ap .........................................(d)

(c) - (d)

(x^2 / a) - 4y = 4ap^2 -4ap^2 -4ap + 4ap + a

(x^2 / a) - 4y = a

(x^2 / a) = 4y + a

x^2 = 4ay + a^2

Therefore x^2 = 4a (y + a/4)


Hope that helps and good Luck