Exercise 10.4 answers: Conic Sections

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1. The equation is already in the standard form x²/a² − y²/b² = 1, so a² = 16 and b² = 9.
2. The transverse axis is along the x-axis, so the vertices are (± a, 0) = (± 4, 0).
3. c² = a² + b² = 16 + 9 = 25, so c = 5 and the foci are (± 5, 0).
4. Eccentricity e = c/a = 5/4.
5. Length of the latus rectum = 2b²/a = 18/4 = 9/2.

AnswerFoci (± 5, 0); vertices (± 4, 0); eccentricity 5/4; latus rectum 9/2.

Watch the lesson Once c outgrows a, the ellipse's eccentricity and equation become the hyperbola's

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1. y² carries the plus sign, so this is the standard form y²/a² − x²/b² = 1 with a² = 9, b² = 27.
2. The transverse axis is along the y-axis, so the vertices are (0, ± a) = (0, ± 3).
3. c² = a² + b² = 9 + 27 = 36, so c = 6 and the foci are (0, ± 6).
4. Eccentricity e = c/a = 6/3 = 2.
5. Length of the latus rectum = 2b²/a = 54/3 = 18.

AnswerFoci (0, ± 6); vertices (0, ± 3); eccentricity 2; latus rectum 18.

Watch the lesson Once c outgrows a, the ellipse's eccentricity and equation become the hyperbola's

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1. Divide throughout by 36 to get the standard form: y²/4 − x²/9 = 1.
2. So a² = 4 and b² = 9, with the transverse axis along the y-axis.
3. Vertices are (0, ± a) = (0, ± 2).
4. c² = a² + b² = 4 + 9 = 13, so c = √13 and the foci are (0, ± √13).
5. Eccentricity e = c/a = √13/2. Latus rectum = 2b²/a = 18/2 = 9.

AnswerFoci (0, ± √13); vertices (0, ± 2); eccentricity √13/2; latus rectum 9.

Watch the lesson Once c outgrows a, the ellipse's eccentricity and equation become the hyperbola's

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1. Divide throughout by 576 to get the standard form: x²/36 − y²/64 = 1.
2. So a² = 36 and b² = 64, with the transverse axis along the x-axis.
3. Vertices are (± a, 0) = (± 6, 0).
4. c² = a² + b² = 36 + 64 = 100, so c = 10 and the foci are (± 10, 0).
5. Eccentricity e = c/a = 10/6 = 5/3. Latus rectum = 2b²/a = 128/6 = 64/3.

AnswerFoci (± 10, 0); vertices (± 6, 0); eccentricity 5/3; latus rectum 64/3.

Watch the lesson Once c outgrows a, the ellipse's eccentricity and equation become the hyperbola's

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1. Divide throughout by 36 to get the standard form: y²/(36/5) − x²/4 = 1.
2. So a² = 36/5 and b² = 4, with the transverse axis along the y-axis.
3. Vertices are (0, ± a) = (0, ± 6/√5) = (0, ± 6√5/5).
4. c² = a² + b² = 36/5 + 4 = 56/5, so c = 2√70/5 and the foci are (0, ± 2√70/5).
5. Eccentricity e = c/a = √14/3. Latus rectum = 2b²/a = 4√5/3.

AnswerFoci (0, ± 2√70/5); vertices (0, ± 6√5/5); eccentricity √14/3; latus rectum 4√5/3.

Watch the lesson Once c outgrows a, the ellipse's eccentricity and equation become the hyperbola's

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1. Divide throughout by 784 to get the standard form: y²/16 − x²/49 = 1.
2. So a² = 16 and b² = 49, with the transverse axis along the y-axis.
3. Vertices are (0, ± a) = (0, ± 4).
4. c² = a² + b² = 16 + 49 = 65, so c = √65 and the foci are (0, ± √65).
5. Eccentricity e = c/a = √65/4. Latus rectum = 2b²/a = 98/4 = 49/2.

AnswerFoci (0, ± √65); vertices (0, ± 4); eccentricity √65/4; latus rectum 49/2.

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1. Vertices and foci are on the x-axis, so the transverse axis lies along the x-axis.
2. Vertices (± 2, 0) give a = 2.
3. Foci (± 3, 0) give c = 3.
4. For a hyperbola, b² = c² − a² = 9 − 4 = 5.
5. Write the standard equation x²/a² − y²/b² = 1 with a² = 4, b² = 5.

Answerx²/4 − y²/5 = 1

Watch this explained “Where the sign enters”, 4:33 into Once c outgrows a, the ellipse's eccentricity and equation become the hyperbola's

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1. The foci are on the y-axis, so the equation has the form y²/a² − x²/b² = 1.
2. The vertices are (0, ± 5), so a = 5, giving a² = 25.
3. The foci are (0, ± 8), so c = 8.
4. For a hyperbola, b² = c² − a² = 64 − 25 = 39.
5. Put a² = 25 and b² = 39 into the form.

Answery²/25 − x²/39 = 1

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1. The foci are on the y-axis, so the equation has the form y²/a² − x²/b² = 1.
2. The vertices are (0, ± 3), so a = 3, giving a² = 9.
3. The foci are (0, ± 5), so c = 5.
4. b² = c² − a² = 25 − 9 = 16.

Answery²/9 − x²/16 = 1

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1. The foci are on the x-axis, so the equation has the form x²/a² − y²/b² = 1.
2. Transverse axis = 2a = 8, so a = 4, giving a² = 16.
3. The foci are (± 5, 0), so c = 5.
4. b² = c² − a² = 25 − 16 = 9.

Answerx²/16 − y²/9 = 1

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1. The foci are on the y-axis, so the equation has the form y²/a² − x²/b² = 1.
2. Conjugate axis = 2b = 24, so b = 12, giving b² = 144.
3. The foci are (0, ± 13), so c = 13.
4. a² = c² − b² = 169 − 144 = 25.

Answery²/25 − x²/144 = 1

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1. The foci are on the x-axis, so the equation has the form x²/a² − y²/b² = 1.
2. c² = (3√5)² = 45.
3. Latus rectum = 2b²/a = 8, so b² = 4a.
4. Using c² = a² + b²: 45 = a² + 4a.
5. Solving a² + 4a − 45 = 0 gives a = 5 (the negative root is rejected).
6. b² = 4 × 5 = 20.

Answerx²/25 − y²/20 = 1

Watch this explained “Reading it off, and working it back”, 12:37 into The latus rectum measured on an open curve, by the ellipse's own calculation

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1. The equation has the form x²/a² − y²/b² = 1.
2. c² = 16.
3. Latus rectum = 2b²/a = 12, so b² = 6a.
4. Using c² = a² + b²: 16 = a² + 6a.
5. Solving a² + 6a − 16 = 0 gives a = 2 (the negative root is rejected).
6. b² = 6 × 2 = 12.

Answerx²/4 − y²/12 = 1

Watch this explained “Reading it off, and working it back”, 12:37 into The latus rectum measured on an open curve, by the ellipse's own calculation

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1. The equation has the form x²/a² − y²/b² = 1.
2. a = 7, so a² = 49.
3. e = c/a, so c = a × e = 7 × 4/3 = 28/3.
4. b² = c² − a² = 784/9 − 441/9 = 343/9.

Answerx²/49 − 9y²/343 = 1

Watch this explained “The same ratio”, 0:53 into Once c outgrows a, the ellipse's eccentricity and equation become the hyperbola's

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1. For a hyperbola with foci on the y-axis: y²/a² − x²/b² = 1.
2. Given: c = √10, so c² = 10.
3. From c² = a² + b²: a² + b² = 10.
4. The hyperbola passes through (2, 3): 9/a² − 4/b² = 1.
5. Let u = a², v = b². Then u + v = 10 and 9/u − 4/v = 1.
6. Substitute v = 10 − u: 9/u − 4/(10−u) = 1.
7. Expand: 9(10−u) − 4u = u(10−u).
8. Simplify: 90 − 9u − 4u = 10u − u².
9. Rearrange: u² − 23u + 90 = 0.
10. Factor: (u − 18)(u − 5) = 0.
11. Since u = 18 gives v = −8 (invalid), we have u = 5 and v = 5.
12. Therefore a² = 5 and b² = 5.

Answery²/5 − x²/5 = 1

Watch this explained “The other way up”, 12:36 into Once c outgrows a, the ellipse's eccentricity and equation become the hyperbola's