Solving Parabola given the equation?

Question

find the vertex and rectum.

Answer 1

sure, once you give the parabola,
what is the rectum of a parabola, anyhow?.

Answer 2

Without more detail, it looks like you want the derivation. Your parabola could open up or down, left or right, but I'll do it for the standard form

y = ax^2 + bx + c, with a>0 (Equation 1)

You can change signs to make it open downward, or interchange x and y to make it open horizontally.

To get this, you must understand the geometry. For this parabola, the "vertex" is the minimum point; we'll call it (h,k). There's also a point, called the "focus," located some distance d directly above the vertex along the "line of symmetry" at (h,k+d). The "latus rectum" (a Latin name) has length 4d, and the parabola passes through the points (h-2d,k+d) and (h+2d,k+d).

Get out a sheet of paper, draw the x and y axes, locate the vertex somewhere -- I'd use the first quadrant -- at (h,k). Pick some small value d, and plot the three points (h,k+d), (h-2d,k+d), and (h+2d,k+d). Draw the parabola passing through the vertex and the two ends of the latus rectum. Now that you can see the geometry, let's do the algebra.

In a minute, we're going to get Equation 1 into this form:

4d (y - k) = (x - h)^2 (Equation 2)

But first, satisfy yourself that in Equation 2, if x=h, then y=k, and the point (h,k) is the minimum of the y function, i.e., the vertex.

Now let x = h +/- 2d. Equation 2 becomes

4d (y - k) = (h +/- 2d - h)^2 = 4d^2
y - k = d
y = k + d

so for x = h+/-2d, y = k+d, and the parabola passes through both ends of the latus rectum.

Finally, we'll do the derivation. Working backwards, we'll start with Equation 2:

4d (y - k) = (x - h)^2

4dy - 4dk = x^2 - 2hx + h^2

4dy = x^2 - 2hx + h^2 + 4dk

y = (1/4d)x^2 - (h/2d)x + h^2 / 4d + k (Equation 3)

Now we set Equation 3 equal to Equation 1:

a = 1/4d ==> d = 1/4a (the focus)
b = -h/2d ==> h = -2db = -b/2a
c = h^2 / 4d + k
k = c - (b^2 / 4a^2)(a) = c - b^2/4a

So given the equation of a parabola in the form of Equation 1, the vertex (h,k) is at (-b/2a, c-b^2/4a), the focus (h,k+d) is at (-b/2a, c-(b^2-1)/4a), and the length of the latus rectum is four times the focus, or 1/a. (ANSWER)


Having done this much, let's construct an example. Suppose we want the vertex to be at (5,3) with a focal length of 2 and a latus rectum of 8. Then the parabola passes through (1,5) and (9,5).

We have three points that must satisfy Equation 1:

ax^2 + bx + c = y
(1) For (1,5) a + b + c = 5
(2) For (5,3) 25a + 5b + c = 3
(3) For (9,5) 81a + 9b + c = 5

Solving 3 equations in 3 unknowns:

(4=3-2): 56a + 4b = 2
(5=2-1): 24a + 4b = -2
(6=4-5): 32a = 4 ==> a = 1/8 (a coefficient)
From (5): 24/8 +4b = -2 ==> b = -5/4 (another coefficient)
From (1): 1/8 - 5/4 + c = 5 ==> c = 49/8 (third coefficient)

The equation of the parabola is

y = (1/8)x^2 - (5/4)x + 49/8 (Equation 4), or

8y = x^2 - 10x + 49

You can satisfy yourself that this parabola passes through (1,5), (5,3), and (9,5).

From our results above, and using Equation 4 where a=1/8, b=-5/4, and c=49/8, the vertex is at

x = h = -b/2a = -(-5/4)/(2/8) = 5
y = k = c - b^2/4a = 49/8 - (25/16)(2) = 24/8 = 3

The focal length d = 1/4a = 2, and the length of the latus rectum is 4d = 8.

So the vertex is at (5,3); the focus is at (5,5); and the ends of the latus rectum are at (1,5) and (9,5).

This was not an easy problem to do.