1. Some examples
Example 1. The parabola y = x2 + 3x + 4 opens upward. For which value of x does
y attain its smallest value?
We complete the square and rewrite the equation as
y = x2 + 3x + ( 32 ) − ( 32 ) + 4
2
2
= (x + 32 ) + 4 − 49 .
So we have that y = (x + 32 )2 + 74 . This implies that the minimum value is y =
and occurs when x = − 23 .
2
7
4
Example 2. Cut the parabola y = x2 + 3x + 4 by the line y = 4. Compute the area
of the resulting parabolic section.
We will use Archimedes’ theorem, which is proved in homework 3. Since y = 4 is
horizontal, the vertex of the parabolic section is the vertex of the parabola, (− 32 , 74 ).
We need the points of intersection for x2 + 3x + 4 = 4. This gives x2 + 3x = 0, or
x(x + 3) = 0. SO x = 0, −3. So our points of intersection are (0, 4) and (−3, 4).
Now consider the triangle formed by the points (−3, 4), (− 23 , 74 ), and (0, 4). This
gives a triangle with base 3 and height 49 . So it has area 12 ⋅3⋅ 94 = 27
8 . By Archimedes’
4 27
theorem, we have that the area of the parabolic section is 3 ⋅ 8 = 92 .
Example 3. Suppose P = (x, y) is a point in the plane. What is the relationship
between x and y that puts P into the region in the previous example?
To be in the region, the point has to be above the parabola y = x2 + 3x + 4 and
below the line y = 4. This happens when x2 + 3x + 4 ≤ y ≤ 4. So we could think of
the parabolic section as the set of points {(x, y)∣x2 + 3x + 4 ≤ y ≤ 4}.
2. Circles
Let’s return to that all-important shape, the circle. We’ve seen circles from the
point of view of the Greeks. We’ve also seen them in the context of ellipses whose
focal points at in the same place. Now that we have the Cartesian plane, let’s
consider the equation of a circle. For example, consider a circle with center at the
point (3, 2) and with a radius of 4. The the circle is the set of points P that are
exactly distance 4 from the point (3,2). For P = (x, y), this happens when,
√
(x − 3)2 + (y − 2)2 = 4.
1
2
Or when,
(x − 3)2 + (y − 2)2 = 16.
See figure 4.18 in your text for reference.
Of course there was nothing special about the point (3,2) or a radius of 4. Suppose we have a center(h, k) and a radius r. Then the equation for the corresponding
circle is
(x − h)2 + (y − k)2 = r2 .
Of course, when we center the circle at the origen, we have the equation as
x2 + y 2 = r 2 .
Example 4. The equation of the circle with center (2, -5) and radius 3 is
(x − 2)2 + (y + 5)2 = 9.
Example 5. Show that x2 + y 2 + 2x − 6y + 7 = 0 is the equation for a circle.
Rewrite the equation as
(x2 + 2x) + (y 2 − 6y) = −7.
We will now complete the squares for each variable to get
(x2 + 2x + 1) + (y 2 − 6y + 9) = −7 + 1 + 9.
That is,
(x + 1)2 + (y − 3)2 = 3.
√
This is an equation for the circle wit radius 3 and center (-1, 3).