Parabolic Minimums

Today's question deals with parabolas, lines, and some calculus. It's not very hard, but, as always, it's very interesting.

Two parabolas, y = x²/x + 7 and y = -x²/4 + 3x, are drawn on the same coordinate plane. A vertical line is going through the parabolas. P and Q are the points of intersections with the first and second parabolas, respectively. Find the vertical line and the shortest distance between P and Q.

The first thing to do when you face such a question is to find an equation that represents the quantity you want, this time the distance PQ. Since the two points are on the same vertical line, they both have the same x-coordinate. Let's call it x. That makes the points:
P(x, x²/2 + 7)
Q(x, -x²/4 + 3x)

The distance between them is just the difference in the y-coordinates (because they lie on a vertical line). The distance is:
d = x² + 7 - (-x²/4 + 3x) = x²/2 + 7 + x²/4 - 3x = 3x²/4 - 3x + 7

We want the shortest d possible, and that's achieved when the derivative of d is zero:
d' = 3x/2 - 3
0 = 3x/2 - 3
3 = 3x/2
6 = 3x
x = 2

Now we know the vertical line, x = 2. Plug this value in d and find the shortest distance between P and Q:
d = 3 * 2²/4 - 3 * 2 + 7 = 3 - 6 + 7 = 4

The shortest distance between P and Q is 4.

Have a great week,
Nadav

nadavs

Parametric Parabolas

It hasn't even been a week since we dealt with parabolas, and they come again, this time with actual numbers. The question itself isn't hard, but once again, some thinking is required.

A parabola goes through the points A(17, 3) B(34, z) C(50, -30). B is the maximum point of the parabola. Find point B and the parabola function in standard form.

As you know, a parabola can be written in two ways. One is standard form, with the a, b, and c coefficients, and one is the vertex form. To solve this question, we'll use the vertex form. This form says that the equation a parabola that has a vertex (h, k) is:
y = a(x - h)² + k

We know h and k, so let's plug them in:
y = a(x - 34)² + z

We have two other points, so let's use them:
3 = a(17 - 34)² + z
2 = a(50 - 34)² + z

3 = 289a + z
-30 = 256a + z

Subtract the second equation from the first:
33 = 33a
a = 1
3 = 289a + z
3 = 289 + z
z = -286

The function of the parabola is:
y = (x - 34)² - 286

Open parenthese and combine like terms to switch to standard form:
y = x² - 68x + 1156 - 286
y = x² - 68x + 870

Nice, isn't it?
Nadav

nadavs

Parabola Reflections

Today's question is of a type we didn't have for a long time - pure variables, no numbers. It's very interesting, and although it looks impossible at first, one little step makes it very, very easy.

The parabola y = ax² + bx + c has its vertex on (h, k). The parabola is reflected about the line y = k. The new parabola is: y = dx² + ex + f. Find a + b + c + d + e + f.

Looks impossible, doesn't it? You can try doing some complex calculations showing that -b/2a = h (you should know that), then plug it back, and equate it to y. That's hard. However, parabolas have two forms: standard form and vertex form. Guess what - we have the vertex!

y = a(x - h)² + k

Now we can open the parentheses and find exactly a, b, and c:
y = ax² - 2axh + ah² + k

That means:
a = a
b = -2ah
c = ah² + k

When the line is reflected about y = k, it changes the coordinates. Since the vertex is on y = k, the parabola just switches sides. This means a = -a and b becomes the opposite too (2ah) to keep the vertex.

Notice that the vertex remains the same because the line was reflected about y = k. When you reflect point (x, n) about the line y = z, you end up with (x, 2z - n). If you reflect (h, k) about y = k, you get (h, 2k - k), which is (h, k).

Since each point turns into (x, 2k - y), the new y-intercept, f, becomes (0, 2k - (ah² + k)), which is (0, 2k - ah² - k), or (0, k - ah²).

a and d are opposites, so a + d = 0. b and e behave the same. The only addends left are c and f:
c + f = ah² + k + k - ah² = 2k

This big sum, a + b + c + d + e + f, equals to 2k.

Hope you liked it,
Nadav

nadavs