1. STANDARD FORM OF EQUATIONS OF TRANSLATED CONICS
WRITING AND GRAPHING EQUATIONS OF CONICS
GRAPHS OF RATIONAL FUNCTIONS
STANDARD FORM OF EQUATIONS OF TRANSLATED CONICS
In the following equations the point (h, k) is the vertex of the parabola and the center of the other conics.
CIRCLE
(x – h) 2 + (y – k) 2 = r 2
Horizontal axis
Vertical axis
PARABOLA
(y – k) 2 = 4p (x – h)
(x – h) 2 = 4p (y – k)
ELLIPSE
(x – h) (y – k) 2
= 1
a 2
b 2
HYPERBOLA
(x – h) (y – k) 2
– = 1
b 2
a 2
(y – k) (x – h) 2
– = 1
b 2
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2. Naserellid (–2, 1) Write an equation of the parabola
Writing an Equation of a Translated Parabola
Write an equation of the parabola
whose vertex is at (–2, 1)
and whose focus is at (–3, 1).
(–2, 1)
SOLUTION
Choose form: Begin by sketching the
parabola. Because the parabola opens
to the left, it has the form :
where p < 0.
(y – k) 2 = 4p(x – h)
Find h and k: The vertex is at (–2, 1),
so h = – 2 and k = 1.
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3. The standard form of the equation is:
Writing an Equation of a Translated Parabola
Write an equation of the parabola
whose vertex is at (–2, 1)
and whose focus is at (–3, 1).
(–2, 1)
SOLUTION
(–3, 1)
Find p: The distance between the vertex (–2, 1), and the focus (–3, 1) is
p = (–3 – (–2)) 2 + (1 – 1) 2 = 1
so p = 1 or p = – 1. Since p < 0, p = – 1.
The standard form of the equation is:
(y – 1) 2 = – 4(x + 2).
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4. Naserellid (x – h) 2 + (y – k) 2 = r 2 (h, k) = (3, – 2). SOLUTION
Graphing the Equation of a Translated Circle
Graph (x – 3) 2 + (y + 2) 2 = 16.
SOLUTION
Compare the given equation to the
standard form of the equation of a circle:
(x – h) 2 + (y – k) 2 = r 2
(3, – 2)
You can see that the graph will be a circle with center at
(h, k) = (3, – 2).
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5. Naserellid SOLUTION (3, 2) r (3 + 4, – 2 + 0) = (7, – 2) (– 1, – 2)
Graphing the Equation of a Translated Circle
Graph (x – 3) 2 + (y + 2) 2 = 16.
SOLUTION
(3, 2)
The radius is r = 4
Plot several points that are each
4 units from the center: r
(3 + 4, – 2 + 0) = (7, – 2)
(– 1, – 2)
(3, – 2)
(7, – 2)
(3 – 4, – 2 + 0) = (– 1, – 2)
(3 + 0, – 2 + 4) = (3, 2)
(3, – 6)
(3 + 0, – 2 – 4) = (3, – 6)
Draw a circle through the points.
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6. Naserellid Write an equation of the ellipse with
Writing an Equation of a Translated Ellipse
Write an equation of the ellipse with
foci at (3, 5) and (3, –1) and vertices at (3, 6) and (3, –2).
(3, 5)
(3, –1)
(3, 6)
(3, –2)
SOLUTION
Plot the given points and make a rough sketch.
(x – h) (y – k) 2
= 1
a 2
b 2
The ellipse has a vertical major axis,
so its equation is of the form:
Find the center: The center is halfway between the vertices.
(3 + 3) ( –2) 2
(h, k) = , = (3, 2)
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7. Naserellid Write an equation of the ellipse with
Writing an Equation of a Translated Ellipse
Write an equation of the ellipse with
foci at (3, 5) and (3, –1) and vertices at (3, 6) and (3, –2).
(3, 5)
(3, –1)
(3, 6)
(3, –2)
SOLUTION
Find a: The value of a is the distance between the vertex and the center.
a = (3 – 3) 2 + (6 – 2) 2 = = 4
Find c: The value of c is the distance between the focus and the center.
c = (3 – 3) 2 + (5 – 2) 2 = = 3
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8. Naserellid The standard form is :
Writing an Equation of a Translated Ellipse
Write an equation of the ellipse with
foci at (3, 5) and (3, –1) and vertices at (3, 6) and (3, –2).
(3, 5)
(3, –1)
(3, 6)
(3, –2)
SOLUTION
Find b: Substitute the values of a and c
into the equation b 2 = a 2 – c 2 .
b 2 = 4 2 – 3 2
b 2 = 7
b = 7 16 7
= 1
The standard form is :
(x – 3) (y – 2) 2
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9. Naserellid (x + 1) 2 SOLUTION 4
Writing an Equation of a Translated Ellipse
Graph (y + 1) 2 – = 1.
(x + 1) 2 4
SOLUTION
The y 2-term is positive, so the transverse axis is vertical. Since a 2 = 1 and b 2 = 4, you know that a = 1 and b = 2.
(–1, –2)
(–1, 0)
(–1, –1)
Plot the center at (h, k) = (–1, –1).
Plot the vertices 1 unit above and below the center at
(–1, 0) and (–1, –2).
Draw a rectangle that is centered at
(–1, –1) and is 2a = 2 units high
and 2b = 4 units wide.
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10. Naserellid (x + 1) 2 SOLUTION 4 (–1, 0) (–1, –1) (–1, –2)
Writing an Equation of a Translated Ellipse
Graph (y + 1) 2 – = 1.
(x + 1) 2 4
SOLUTION
(–1, –2)
(–1, 0)
(–1, –1)
Draw the asymptotes through the corners of the rectangle.
Draw the hyperbola so that it passes
through the vertices and approaches
the asymptotes.
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11. Naserellid The expression B 2 – 4AC is called the discriminate
CLASSIFYING A CONIC FROM ITS EQUATION
The equation of any conic can be written in the form
Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0
which is called a general second-degree equation in x and y.
The expression B 2 – 4AC is called the discriminate
of the equation and can be used to determine which type of conic the equation represents.
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12. Naserellid CLASSIFYING A CONIC FROM ITS EQUATION
CONCEPT
SUMMARY
CONIC TYPES
The type of conic can be determined as follows:
DISCRIMINATE
(B 2 – 4AC)
TYPE OF CONIC
< 0, B = 0, and A = C
Circle
< 0, and either B  0, or A  C
Ellipse
= 0
Parabola
> 0
Hyperbola
If B = 0, each axis is horizontal or vertical. If B  0, the axes are neither horizontal nor vertical.
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13. the value of the discriminate is:
Classifying a Conic
Classify the conic
2 x 2 + y 2 – 4 x – 4 = 0.
Classify the conic
4 x 2 – 9 y x – 144 y – = 0.
SOLUTION
SOLUTION
Since A = 4, B = 0, and C = –9,
the value of the discriminate is:
Since A = 2, B = 0, and C = 1,
the value of the discriminate is:
B 2 – 4 AC = 0 2 – 4 (2) (1) = – 8
B 2 – 4 AC = 0 2 – 4 (4) (–9) = 144
Because B 2 – 4 AC < 0 and A  C,
Because B 2 – 4 AC > 0,.
the graph is an ellipse.
, the graph is a hyperbola.
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14. WRITING AND GRAPHING EQUATIONS OF CONICS
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