#$&* course MTH 164 SOLUTIONS/COMMENTARY ON QUERY 13
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20:43:50 20:43:55
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**** Query problem 9.4.12 graph of hyperbola with vertices at (-4,0) and (4,0), asymptote y = 2x **** Give the equation of the specified hyperbola.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The equation of a hyperbola centered at the origin is x^2 / a^2 - y^2 / b^2 = 1, where (a, 0) and (-a, 0) are the vertices and y = +-b / a * x are the asymptotes. In the present case we have a = 4 and y = 2 x is an asymptote. It follows that b / a = 2 so that b = 2 a = 2 * 4 = 8. Thus the equation is x^2 / 4^2 - y^2 / 8^2 = 1 or x^2 / 16 - y^2 / 64 = 1. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The equation of a hyperbola centered at the origin is x^2 / a^2 - y^2 / b^2 = 1, where (a, 0) and (-a, 0) are the vertices and y = +-b / a * x are the asymptotes. In the present case we have a = 4 and y = 2 x is an asymptote. It follows that b / a = 2 so that b = 2 a = 2 * 4 = 8. Thus the equation is x^2 / 4^2 - y^2 / 8^2 = 1 or x^2 / 16 - y^2 / 64 = 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:2 Explain how you know whether the equation is x^2 / a^2 - y^2 / b^2 = 1, x^2 / a^2 + y^2 / b^2 = 1 or -x^2 / a^2 + y^2 / b^2 = 1.
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13:18:39 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The point (4, 0) wouldn't make sense in an equation of the form -x^2 / a^2 + y^2 / b^2 = 1, since for this point -x^2 / a^2 would be negative and y^2 / b^2 would be positive, making the left-hand side negative while the right-hand side is 1. For large x and y the value of x^2 / a^2 + y^2 / b^2 would be large and couldn't be equal to 1. The form x^2 / a^2 - y^2 / b^2 = 1 makes sense for (4, 0) and also for large x and y, since the difference x^2 / a^2 - y^2 / b^2 could indeed equal 1 even for large x and y confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The point (4, 0) wouldn't make sense in an equation of the form -x^2 / a^2 + y^2 / b^2 = 1, since for this point -x^2 / a^2 would be negative and y^2 / b^2 would be positive, making the left-hand side negative while the right-hand side is 1. For large x and y the value of x^2 / a^2 + y^2 / b^2 would be large and couldn't be equal to 1. The form x^2 / a^2 - y^2 / b^2 = 1 makes sense for (4, 0) and also for large x and y, since the difference x^2 / a^2 - y^2 / b^2 could indeed equal 1 even for large x and y. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:2 **** Query problem 9.4.26 graph of hyperbola defined by rectangle of width 4, centered at the origin, lines y = + - 2x passing through diagonals, opening right and left. **** Give the equation of the specified hyperbola.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The width of the rectangle is 4 so its vertical sides are at x = -2 and x = 2. Since the parabola opens to the right the vertices are on the right and left sides of the rectangle, at (-2, 0) and (2, 0). The lines y = +- 2 x are the asymptotes. The form of the equation is x^2 / a^2 - y^2 / b^2 = 1, with vertices (a, 0) and (-a, 0) and asymptotes y = +- b / a * x. It follows that a = 2 (vertices at (-a,0) and (a,0)) and b / a = 2 (since y = +-2 x = +- b / a * x). We get b = 2 a = 2 * 2 = 4 so the equation is x^2 / 2^2 - y^2 / 4^2 = 1 or x^2 / 4 - y^2 / 16 = 1. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The width of the rectangle is 4 so its vertical sides are at x = -2 and x = 2. Since the parabola opens to the right the vertices are on the right and left sides of the rectangle, at (-2, 0) and (2, 0). The lines y = +- 2 x are the asymptotes. The form of the equation is x^2 / a^2 - y^2 / b^2 = 1, with vertices (a, 0) and (-a, 0) and asymptotes y = +- b / a * x. It follows that a = 2 (vertices at (-a,0) and (a,0)) and b / a = 2 (since y = +-2 x = +- b / a * x). We get b = 2 a = 2 * 2 = 4 so the equation is x^2 / 2^2 - y^2 / 4^2 = 1 or x^2 / 4 - y^2 / 16 = 1. ** 20:59:30 20:59:40
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:2 **** Query problem 9.6.6 conic defined by a r = 6 / (8 + 2 sin(`theta)). **** What conic is defined by the given polar equation?
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21:00:25 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since e < 1 the conic is an elipse. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since e < 1 the conic is an elipse. ** Good. For reference two derivations are included below: ** ** Derivation of forms of a conic: If the directrix D oriented perpendicular to the polar axis lies at distance p in the negative direction from that axis, then the distance from point P = (r, `theta) to the directrix will be measured in the direction parallel to the polar axis. r cos(`theta) is the displacement of P with respect to the polar axis do this distance is dist(D,P) = r cos(`theta) + p. If the focus is at the pole then dist(F,P) = distance from pole to P = r. The ratio of the latter distance to the former is the eccentricity so we have: r / [ r cos(`theta) + p ] = e r = e r cos(`theta) + e p r - e r cos(`theta) = e p r = e p / (1 - e cos(`theta) ). For directrix p units above pole parallel to polar axis we have for point P dist(D, P) = r sin(`theta) - p dist(F, P) = r so r / [ r sin(`theta) - p ] = e r = e r sin(`theta) - e p etc. r = e p / (e r sin`theta - 1) ** ** We can convert to rectangular coordinates. We can rearrange the equation to get 8 r + 2 r sin(`theta) = 6. Since r sin(`theta) = y and r = `sqrt(x^2+y^2) we have 8 `sqrt(x^2+y^2) + 2 y = 6. Getting the `sqrt on one side we have `sqrt(x^2 + y^2) = (6-2y)/8. Squaring both sides we have x^2 + y^2 = (36 - 24 y + 4y^2) / 64 so that 64 x^2 + 60 y^2 + 24 y = 36. If we complete the square on y we do end up with a positive number on the right-hand side, so we indeed have an ellipse. **
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21:00:26
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:2 **** How do you know that the conic is the one you indicated?
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21:02:19 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: After dividing both numerator and denominator by 8, I get r = (3/4) / (1+ 1/4sin('theta)) so I know that e = 1/4 which is less than 1. This results in an elipse. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT SOLUTION: After dividing both numerator and denominator by 8, I get r = (3/4) / (1+ 1/4sin('theta)) so I know that e = 1/4 which is less than 1. This results in an elipse.
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21:02:20
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:2 **** What is the position of the directrix, if one exists?
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21:03:05 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The directrix is parallel to the polar axis at a distance of p = 3 above the pole. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT ANSWER: The directrix is parallel to the polar axis at a distance of p = 3 above the pole.
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21:03:06 21:03:13
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:2 **** Query problem 9.6.16 identify and graph r(2-cos(`theta)) = 2. **** Describe in detail the graph you obtain from the given equation.
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21:07:33 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The conic is a parabola which opens to the right, with a vertex at (1/2, 'pi), a directrix that passes through (1, 'pi) that is perpendicular to the polar axis. The focus is at the pole. The parabola also contains the points (1, 'pi/2) and (1, 3'pi/2). confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The conic is a parabola which opens to the right, with a vertex at (1/2, 'pi), a directrix that passes through (1, 'pi) that is perpendicular to the polar axis. The focus is at the pole. The parabola also contains the points (1, 'pi/2) and (1, 3'pi/2). ** Good. The equation rearranges to give you r = 2 / (2 - cos(`theta) ); dividing numerator and denominator by 2 you get r = 1 / (1 - 1/2 cos(`theta) ). So e = 1/2 and since ep = 1/2 p = 1 we have p = 2. **
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21:07:34 21:07:38
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:2 **** Query problem 9.6.28 convert to rectangular coordinates r(2-cos(`theta)) = 2. **** What is the rectangular coordinate form of the given equation, and how did you obtained it a?
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21:12:21 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: r ( 2- cos ('theta)) = 2 2r - r cos('theta) = 2 distribute the r 2r = 2 + r cos('theta) rearrange 4r^2 = ( 2 + r cos('theta))^2 square both sides 4 (x^2 + y^2) = (2 + x)^2 use transformations 4x^2 + 4y^2 = x^2 + 4x +4 distribute 4y^2 = -3x^2 + 4x +4 isolate the y confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: r ( 2- cos ('theta)) = 2 2r - r cos('theta) = 2 distribute the r 2r = 2 + r cos('theta) rearrange 4r^2 = ( 2 + r cos('theta))^2 square both sides 4 (x^2 + y^2) = (2 + x)^2 use transformations 4x^2 + 4y^2 = x^2 + 4x +4 distribute 4y^2 = -3x^2 + 4x +4 isolate the y ** Keeping the equation in the standard rectangular form of a conic we have 3 x^2 + 4 y^2 - 4 x - 4 = 0. From this form it should evident that completing the square on x will give you the standard equation of an ellipse. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:2 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!