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**preview**shows pages 1-2. to view the full**6 pages of the document.**MA 141 Chapter 0

Section 0.2: Conic Sections

Distance Formula: (derived from Pythagorean Theorem) Given two points P= (x1, y1)

and Q= (x2, y2)in R2, the distance between them is

d(P, Q) = p(x1−x2)2+ (y1−y2)2.

Def: Aparabola is the set of points in the xy-plane that are equidistant from a ﬁxed point,

the focus, and a ﬁxed line, the directrix. The vertex is the point of the parabola that

minimizes the distance to the focus and directrix.

D1=D2

The standard equation of the vertical parabola is

y−k=a(x−h)2.

In this case, the focus is h, k +1

4aand the directrix is y=k−1

4a.

If a > 0, the parabola opens upward and the vertex is the lowest point of the graph. If a < 0,

the parabola opens downward and the vertex is the highest point of the graph.

Similarly, we can have a parabola opening left or right. The standard equation in this

case is

x−h=a(y−k)2.

The focus is at h+1

4a, kand the directrix is x=h−1

4a.

If a > 0, the parabola opens to the right and the vertex is the left-most point. If a < 0, the

parabola opens left and the vertex is the right-most point.

1

Only pages 1-2 are available for preview. Some parts have been intentionally blurred.

Ex: Find the focus, vertex, and directrix of x=−1

6y2and then sketch the parabola.

This parabola is of the second type, opening either left or right. Since a=−1

6is negative,

it opens to the left.

This can be rewritten as x−0 = −1

6(y−0)2, so the vertex is at (0,0).

The focus is then given by 1

4a,0.

1

4a=1

4(−1/6) =−1

4/6=−6

4=−3

2

Then the focus is −3

2,0and the directrix is x=−−3

2=3

2.

−=

Note that if the parabola is not in the given form above, we can complete the square to

convert to the standard equation.

Recall: Given a quadratic y=x2+ 8x+ 3, we complete the square as follows.

y=x2+ 8x+ 3

y+8

22

= x2+ 8x+8

22!+ 3

y+ 16 = (x2+ 8x+ 16) + 3

y+ 13 = (x+ 4)2

Then the standard form of the parabola is y+ 13 = (x+ 4)2.

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