Center ( - 2, 1 ), a = 4, b = 2,

Foci = 
Vertices = ( - 2, 1 ± 4 ) = ( - 2, 5 ) and ( - 2, - 3 )

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Exercise 9.3.5

The distance from the center to a vertex is 1, so a = 1.
The distance from the center to a focus is 2, so c = 2.
Since the square of c equals the sums of the squares of a and b, b must equal the square root of 3.

Since a focus lies below the center, this is a vertical hyperbola.

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Exercise 9.3.4

The distance from the center to a vertex is 3, so a = 3.
The distance from the center to a focus is 5, so c = 5.
Since the square of c equals the sums of the squares of a and b, b = 4.
Since the center, vertices and foci lie on the same horizontal line, this is a horizontal hyperbola.

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Exercise 9.3.3

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Exercise 9.3.2

The distance from V₁ to F₁ equals the distance from V₂ to to F₂.
The distance from V₁ to F₂ equals 2a plus the distance from V₂ to F₂,
which is the same as 2a plus the distance form V₁ to F₁.
So the distance from V₁ to F₂ minus the distance from V₁ to F₁ equals 2a.

Likewise, the distance from V₂ to F₁ minus the distance from V₂ to F₂ equals 2a.

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Exercise 9.3.1

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Exercise 9.2.8

Center ( 1, -1 ), vertices ( 1, -1 ± 5 ), ( 1 ± 3, -1 ), or ( 1, 4 ), ( 1, - 6 ), ( 4, - 1 ), ( - 2, -1 )

Since this is a vertical ellipse, the foci are ( 1, - 1 ± 4 ) = ( 1, 3 ) and ( 1, - 5 ).

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Exercise 9.2.7

A major vertex lies 3 units below the center, so a = 3. A minor vertex lies 1 unit to the right of the center, so b = 1, and the ellipse is vertical. Thus

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Exercise 9.2.6

The focus is two units above the center, so c = 2 and the ellipse is vertical. A major vertex is four units below the center, so a = 4. Since the square of c equals the difference between the squares of a and b, it follows that b equals the square root of 12, or twice the square root of 3. In any event, the square of b is 12. Thus the equation is

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Exercise 9.2.5

The distance from the center to the focus is 1, so c = 1.
The distance from the center to a major vertex is 7, so a = 7.
The square of c equals the square of a minus the square of b, so the square of b equals 48. So b equals four times the square root of three.

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Exercise 9.2.4

The distance from the center to a focus is 3, so c = 3.
The distance from the center to a minor vertex is 4, so b = 4.
The square of c equals the square of a minus the square of b, so the square of a is 25. So a equals five.

Since a minor axis lies above the center, the ellipse is horizontal.

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Exercise 9.2.3

Since the sum of the distances from V₃ to F₁ and V₃ to F2 is 2a and since the distance from V₃ to F₁ equals the distance from V₃ to F2, the distance from V₃ to F2 equals a. Since a is the hypotenuse of a right triangle with sides b and c, then the realtionship between a, b and c is determined by the Pythagorean Theorem.

a² = b² + c²

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Exercise 9.2.2

The distance from V₁ to F₁ equals the distance from F₂ to V₂, so the sum of the distance of V₁ to F₁ plus the distance from V₁ to F₂ equals the sum of the distances from V₁ to F₂ plus the distance from V₂ to F₂, which is just the distance from V₁ to V₂, which is a.

The distance from F₁ to V₃ and from F₁ to V₄ is the same as the distance from the center to V₁, which equals a.

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Exercise 9.2.1

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Exercise 9.1.5

y² + 4 y = 12 x - 40

y² + 4 y + 4 = 12 x - 36

( y + 2 )² = 12 ( x - 3 )

The vertex is ( 3, - 2 ). Since 4 p = 12, p = 3. So the focus is 3 units to the right of the vertex at the point ( 6, - 2 ). The endpoints of the focal chord lie 2 p = 6 units above and below the focus, so the endpoints are at ( 6, 4 ) and ( 6, - 8 ). The directrix is the vertical line three units to the left of the vertex, so itís equations is x = 0. The directrix is the y-axis.

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Exercise 9.1.4

The focus lies 2 units below the vertex, so p = - 2 and the parabola is vertical. The endpoints of the focal chord lie | 2 p | units from the focus at right angles to the axis of the parabola, so they are at points ( - 4, - 2 ) and ( 4, - 2 ).

x² = - 8 y

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Exercise 9.1.3

The focus lies two units to the right of the vertex, so p = 2 and the parabola is horizontal. The endpoints of the focal chord lie a distance | 2 p | above and below the focus at points ( 1, 7 ) and
( 1, - 1 ).

( y - 3 )² = 8 ( x + 1 )

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Exercise 9.1.2

The focus lies 2 units above the vertex, so the parabola is vertical and p = 2. The endpoints of the focal chord lie | 2 p | to the left and right of the focus at points ( - 4, 2 ) and ( 4, 2 ).

x² = 8 y

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Exercise 9.1.1

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