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Point reflector line y 6 4 Inverse relation x 6 4 4 6
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Original relation
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Illustration for the solution to Problem 7 in Chap 12
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Inverse of original restricted relation
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x 6 4 2 2 Movable 4 vertical line 6 Original restricted relation 4 6
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Illustration for the solution to Problem 8 in Chap 12
526 Worked-Out Solutions to Exercises: 11-19
y 6 4 2 x 6 4 2 4 6 Movable vertical line
Original restricted relation
4 6 Inverse of original restricted relation
Figure B-5
Illustration for the solution to Problem 9 in Chap 12
9 Figure B-5 shows the graph of the original relation with its domain restricted to the reals smaller than or equal to 2 ( gray curve) and its inverse (black curve) The curve for the inverse relation is a half-hyperbola that opens downward and intersects the y axis at (0, 2) A movable vertical line never intersects the black curve at more than one point Therefore, the inverse relation f 1 (x) = (4x 2 /9 + 4)1/2 is a true function of x 10 Figure B-6 shows the graph of the original relation with its range restricted to the nonnegative reals (pair of gray curves) and the graph of its inverse (pair of black curves) A movable vertical line never intersects the graph of the original restricted relation at more than one point, so it s a true function Whenever the movable vertical line intersects the graph of the inverse, that line crosses both curves On this basis, we know that the inverse of the original restricted relation is not a true function
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Original restricted relation y 6 4 2
2 2 4 6
Inverse of x original restricted relation
Point reflector line
Movable vertical line
Figure B-6
Illustration for the solution to Problem 10 in Chap 12
13
1 If the plane passes through the point where the apexes of the two cones meet, the intersection is a point, a straight line, or a pair of lines that intersect Here s how the situations break down: If the plane is slanted so that we d get a circle or ellipse if the plane didn t pass through the apexes, then we get a single point ( figs B-7A and B) If the plane slants so that we d get a parabola if the plane didn t pass through the apexes, then we get a straight line ( fig B-7C) If the plane slants so that we d get a hyperbola if the plane didn t pass through the apexes, we get a pair of lines that cross ( fig B-7D) 2 When we examine the graph of the ellipse, we see that the lower focus is at (0,0) and the lower vertex is at (0, 2) Therefore, the distance between the lower focus and the lower vertex is 2 units, telling us that the focal length is f =2
528 Worked-Out Solutions to Exercises: 11-19
Double circular cone Double circular cone
Flat plane
Flat plane
Flat plane
Flat plane
Double circular cone
Double circular cone
Figure B-7
Illustrations for the solution to Problem 1 in Chap 13
We can also see that the distance between the lower vertex and the directrix is 4 units, because the lower vertex is at (0, 2) and the directrix passes through (0, 6) Therefore f /e = 4 Combining the above two equations and solving, we get e = 1/2 That s the eccentricity of the ellipse 3 When we solved Problem 2, we determined that f = 2 and e = 1/2 Plugging these values into the formulas relating the parameters of the ellipse, we obtain u = 2 + 2 /(1/2) + y and x 2 + y 2 = [(1/2) u]2
13
These simplify to u=6+y and x 2 + y 2 = u2/4 Let s substitute the quantity (6 + y) for u in the second equation above That gives us x 2 + y 2 = (6 + y)2/4 This is an equation for our ellipse, although it s not in standard form 4 We ve been told that both foci and both vertices lie on the y axis, so x = 0 at the lower vertex and also at the upper vertex If we plug x = 0 into the equation we got in the solution to Problem 3, we ll be left with an equation that tells us the y values at both vertices of the ellipse Here it is y 2 = (6 + y)2/4 Multiplying through by 4, we obtain 4y 2 = (6 + y)2 When we multiply out the squared binomial on the right-hand side, we get 4y 2 = 36 + 12y + y 2 This is a quadratic equation Let s morph it with algebra into the standard form for a quadratic That gives us 3y 2 12y 36 = 0 which factors into (3y + 6)(y 6) = 0 The solutions are therefore y = 2 or y=6 The first solution corresponds to the point (0, 2) on the ellipse The second solution corresponds to (0,6) on the ellipse We already know that the coordinates of the lower vertex are (0, 2), so the coordinates of the upper vertex must be (0,6)
530 Worked-Out Solutions to Exercises: 11-19
Now let s find the coordinates of the upper focus We already know that the lower focus is at (0,0) We also know that the lower vertex is at (0, 2), which is 2 units below the lower focus on the y axis Because all ellipses are symmetrical with respect to the foci, the upper focus must be 2 units below the upper vertex on the y axis Therefore, the coordinates of the upper focus must be (0,4) 5 We can find the coordinates of our ellipse s center by averaging the coordinates of the foci The lower focus is at (0,0), while the upper focus is at (0,4) Therefore, the center must be at (0,2) The length of the vertical semi-axis (let s call it b) is the distance between the center and the lower vertex, as shown in Fig B-8 The center is at (0,2) and the lower vertex is at (0, 2), so the vertical semi-axis is 4 units long (We get the same result if we use the center and the upper vertex) To find the length of the horizontal semi-axis, we must know the coordinates of either the left-most point or the right-most point on the ellipse These are the points where the horizontal line y = 2, which passes through the center of the ellipse, intersects the curve To find these points, we can plug y = 2 into the equation we got for the ellipse when we solved Problem 3 When we make the substitutions, we obtain x 2 + 22 = (6 + 2)2/4
Upper vertex is at (0, 6)
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