Download Conics Review - Michael Cavers

Conics Review
We next review equations of parabolas, circles, ellipses and hyperbolas.
Each equation presented is the standard form for the type of conic.
Mnemonic: In each conic formula presented, the terms ‘x − h’ and ‘y − k’ will always appear.
The point (h, k) will alway represent either the centre or vertex of the particular conic.
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Vertical Parabola
The equation of a vertical parabola is:
(h, k) is the vertex of the parabola.
a is the vertical stretch factor.
If a > 0, the parabola opens upward.
If a < 0, the parabola opens downward.
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Horizontal Parabola
The equation of a horizontal parabola is:
(h, k) is the vertex of the parabola.
a is the horizontal stretch factor.
If a > 0, the parabola opens right.
If a < 0, the parabola opens left.
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Circle
The equation of a circle is:
(h, k) is the centre of the circle.
r is the radius of the circle.
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Ellipse
The equation of an ellipse is:
(h, k) is the centre of the ellipse.
a is the horizontal distance from the centre to the edge of the ellipse. b is the vertical
distance from the centre to the edge.
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Horizontal Hyperbola
The equation of a horizontal hyperbola is:
(h, k) is the centre of the hyperbola.
a, b are the reference box values. The box has a centre of (h, k).
a is the horizontal distance from the centre to the edge of the box.
b is the vertical distance from the centre to the edge of the box.
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Sketching a Horizontal Hyperbola
The dotted lines above are called the asymptotes of the hyperbola.
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Vertical Hyperbola
The equation of a vertical hyperbola is:
(h, k) is the centre of the hyperbola.
a, b are the reference box values. The box has a centre of (h, k).
a is the horizontal distance from the centre to the edge of the box.
b is the vertical distance from the centre to the edge of the box.
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Sketching a Vertical Hyperbola
The dotted lines above are called the asymptotes of the hyperbola.
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Completing the square
Completing the square is an important and useful technique.
It allows us to determine the center of a circle/ellipse or the vertex of a parabola.
It is also used to derive the quadratic formula (or solve quadratic equations).
The main idea behind completing the square is to turn:
ax 2 + bx + c
into
a(x − h)2 + k.
One way to complete the square is to use the following formula:
b 2
b2
ax 2 + bx + c = a x +
− 2 + c.
2a
4a
As this formula is complicated, we outline some easier steps in the next example which use
the factoring formula for perfect square trinomials:
r 2 + 2rs + s 2 = (r + s)2
r 2 − 2rs + s 2 = (r − s)2
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Solving a quadratic by completing the square
Example
Solve 2x 2 + 12x − 32 = 0 by completing the square.
Solution: Because I want to demonstrate what you should do when the a value (in front of x 2 )
is not 1, we will not divide by 2 first (though usually you would).
2x 2 + 12x − 32 = 0
6
→
6
2
Start with original equation.
2x 2 + 12x = 32
Move the number over to the other side.
2(x 2 + 6x) = 32
Factor out the“a” from the ax 2 + bx expression.
=3
→
32 = 9
2(x 2 + 6x + 9) = 32 + 2 · 9
2(x + 3)2 = 50
Take # in front of x, divide by 2, square it.
“Add” to both sides (remember the a value!)
Factor the resulting perfect square trinomial.
You have now completed the square!!
x = 2 or x = −8
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Divide by 2, take square roots, subtract 3.
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Example
Find the center and radius of the circle:
y 2 + x 2 − 12x + 8y + 43 = 0.
Solution:
We need to complete the square twice!
Collect the terms with x together and the terms with y together:
(x 2 − 12x) + (y 2 + 8y ) = −43
Add 36 to both sides for the x term (−12 → −6 → 36).
Add 16 to both sides for the y term (8 → 4 → 16):
(x 2 − 12x + 36) + (y 2 + 8y + 16) = −43 + 36 + 16
Factoring gives:
(x − 6)2 + (y + 4)2 = 32 .
Therefore, the centre of the circle is (6, −4) and the radius is 3.
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Determining the type of conic
An equation of the form
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
gives rise to a graph that can be generated by performing a conic section.
These are parabolas, circles, ellipses, and hyperbolas.
The Bxy term involves conic rotation.
For simplicity, we will omit the Bxy term.
To determine the type of graph we analyze the values of A and C .
I
I
I
I
If
If
If
If
A = C , the graph is a circle.
AC > 0 (and A 6= C ), the graph is an ellipse.
AC = 0, the graph is a parabola.
AC < 0, the graph is a hyperbola.
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Example
What type of conic is 4x 2 − y 2 − 8x + 8 = 0? Write it in standard form.
Solution:
We see that A = 4 and C = −1.
Since AC < 0, the conic is a hyperbola.
We will complete the square. First collect x’s and y ’s:
(4x 2 − 8x) − y 2 = −8
Now we factor out 4 from the x terms.
4(x 2 − 2x) − y 2 = −8
Notice that we don’t need to complete the square for the y terms.
For the x terms, we add 1 (divide −2 by two and square it).
Taking into consideration that the “a” value is 4:
4(x 2 − 2x + 1) − y 2 = −8 + 4 · 1
Factoring gives:
4(x − 1)2 − y 2 = −4
A hyperbola in standard form has ±1 on the right side and a positive x 2 on the left side.
Thus, we must divide by 4:
y2
(x − 1)2 −
= −1
4
Now we can see that the equation represents a vertical hyperbola with centre (1, 0).
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