example 3

EXAMPLE 3 Write an equation of a translated parabola

Write an equation of the parabola whose vertex is at (– 2, 3) and whose focus is at (– 4, 3).

SOLUTION

STEP 1

Determine the form of the equation. Begin by making a rough sketch of the parabola. Because the focus is to the left of the vertex, the parabola opens to the left, and its equation has the form (y – k)2 = 4p(x – h) where p < 0.


STEP 2

Identify h and k. The vertex is at (– 2, 3), so h = – 2 and k = 3.

STEP 3

Find p. The vertex (– 2, 3) and focus (4, 3) both lie on the line y = 3, so the distance between them is | p | = | – 4 – (– 2) | = 2, and thus p = +2. Because p < 0, it follows that p = – 2, so 4p = – 8.


The standard form of the equation is (y – 3)2 = – 8(x + 2).

ANSWER

EXAMPLE 4 Write an equation of a translated ellipse

Write an equation of the ellipse with foci at (1, 2) and (7, 2) and co-vertices at (4, 0) and (4, 4).

SOLUTION

STEP 1

Determine the form of the equation. First sketch the ellipse. The foci lie on the major axis, so the axis is horizontal. The equation has this form:

(x – h)2

a2 +(y – k)2

b2 = 1


STEP 2

Identify h and k by finding the center, which is halfway between the foci (or the co-vertices)

(h, k) = 1 + 7 2 + 22 2 )( , = (4, 2)

STEP 3Find b, the distance between a co-vertex and the center (4, 2), and c, the distance between a focus and the center. Choose the co-vertex (4, 4) and the focus (1, 2): b = | 4 – 2 | = 2 and c = | 1 – 4 | = 3.


STEP 4

Find a. For an ellipse, a2 = b2 + c2 = 22 + 32 = 13, so a =

13

ANSWER

The standard form of the equation is

(x – 4)2

13 +(y – 2)2

4 = 1

EXAMPLE 5 Identify symmetries of conic sections

Identify the line(s) of symmetry for each conic section in Examples 1 – 4.

SOLUTION

For the circle in Example 1, any line through the center (2, – 3) is a line of symmetry.

For the hyperbola in Example 2 x = – 1 and y = 3 are lines of symmetry

EXAMPLE 5 Identify symmetries of conic sections

For the parabola in Example 3, y = 3 is a line of symmetry.

For the ellipse in Example 4, x = 4 and y = 2 are lines of symmetry.

GUIDED PRACTICE for Examples 3, 4 and 5

parabola with vertex at (3, – 1) and focus at (3, 2).

SOLUTION

STEP 1

Determine the form of the equation. Begin by making a rough sketch of the parabola. Because the focus is to the left of the vertex, the parabola opens to the left, and its equation has the form (x – h)2 = 4p(y – k) where p > 0.

5.


STEP 2

Identify h and k. The vertex is at (3,– 1), so h = 3 and k = –1.

STEP 3

Find p. The vertex (3, – 1) and focus (3, 2) both lie on the line x = 3, so the distance between them is | p | = | – 2 – (– 1) | = 3, and thus p = + 3. Because p > 0, it follows that p = 3, so 4p = 12.

The standard form of the equation is (x – 3)2 = 12(y + 1).

ANSWER


Write an equation of the hyperbola with vertices at (– 7,3) and (– 1, 3) and foci at (– 9, 3) and (1, 3).

SOLUTION

STEP 1

Determine the form of the equation. First sketch the horizontal. The foci lie on the x- axis with a horizontal traverse axis. The equation has this form:

(x – h)2

a2

(y – k)2

b2 = 1–

6.

STEP 2

Identify h and k by finding the center, which is halfway between the foci (or the co-vertices)

2(h, k) = – 9 + 1 3 + 3

2 )( , = (– 4, 3)


STEP 3Find b, the distance between a vertex and the center (– 4, 3), and c, the distance between a focus and the center. Choose the co-vertex (–7, 3) and the focus (–9, 3): b = | – 7 –(– 4) | = 3 and c = | –9 – (– 4) | = 5.


STEP 4

Find a. For an ellipse, b2 = c2 + a2 = 52 + 32 = 16, so b = 4

ANSWER

The standard form of the equation is

(x + 4)2

9 –(y – 3)2

16 = 1


7. (x – 5)2

64 + (y)2

16 = 1

Identify the line(s) of symmetry for the conic section.

For the ellipse the lines of symmetry are x = 5 and y = 0.

ANSWER


8. (x + 5)2 = 8(y – 2).

For parabola the lie of symmetry are x = – 5

ANSWER



9. (x – 1)2

49 –(y – 2)2

121 = 1

For horizontal lines of symmetry are x = 1 and y = 2.

ANSWER


example 3

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