MATH 241 FALL 2009 HOMEWORK 3 SOLUTIONS
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1 MATH 41 FALL 009 HOMEWORK 3 SOLUTIONS H3P1 (i) We have the points A : (0, 0), B : (3, 0), and C : (x, y) We now from the distance formula that AC/BC = if and only if x + y (3 x) + y = which is equivalent to x + y = (3 x) + y x + y = 4(3 x) + 4y 3x 4x + 3y = 36 x 8x + y = 1 (x 4) + y = 4 Since this is the equation for a circle in standard form, the set of points C in the plane such that AC/BC = must be the circle with center (4, 0) and radius (ii) As the hint suggests, we choose to use a Cartesian coordinate system such that A : (0, 0) and B : (0, 1), and we let C(x, y) as in part (i) Using the distance formula, we have AC/BC = if and only if x + y x + (1 y) = and so x + y = x + (1 y) (as both sides are nonnegative) x + y = x + (1 y) ( 1)x + ( 1)y y = x + y 1 y = 1 ) x + (y = 1 ) x + (y = 1 ( 1) (as 1) ( 1) Again, the set is a circle The center is as > 0 1 ( ) 0, 1 and the radius is 1,
2 MATH 41 FALL 009 HOMEWORK 3 SOLUTIONS (iii) From part (ii), we have a() = 0, b() = So we obtain the following a() = 0 1 b() = b() = r() = = does not exist More precisely, 1, and r() = 1 =, and b() = = As we see, when 1, the center of the circle moves to or along y-axis, depending on whether < 1 or > 1 At the same time the radius length is becoming infinitely large It is clear from geometry that when = 1, the set of all points C such that AC/BC = 1 is the set of points C such that AC = BC Hence, it is the perpendicular bisector to the segment AB This coincides with our intuition that a line is a circle of infinite radius with the center at infinity Investigating what happens when 0 +, ie, AC is becoming much smaller than BC, we obtain a() = 0; 0 + b() = r() = = 0; 1 = 0 Hence, the circle shrins to point A, which again coincides with our intuition H3P (i) Begin with the point-slope form of the equation of a line: y y 1 = m(x x 1 ) Given slope 3 and the point P : (1, ), we have y = 3(x 1) y = 3x 3 y = 3x 1 Since the line intersects the parabola y = 4 x at points A and B, in order to find the x-coordinates of the points of intersection, we solve the equation The quadratic formula yields 4 x = 3x 1 x + 3x 5 = 0 x = 3 ± 9 (4)(1)( 5) ()(1) = 3 ± 9
3 MATH 41 FALL 009 HOMEWORK 3 SOLUTIONS 3 Denoting the intersection points by A and B, we have x A = 3 9 Therefore x A x B = 9, and x B = Substituting x A and x B into y = 3x 1 yields the corresponding y- coordinates of A and B: y A = 3x A 1 and y B = 3x B 1 Therefore y A y B = 3 x A x B = 3 9, and AB = (x A x B ) + (y A y B ) = x A x B + y A y B = (xa x B ) + 3 (x A x B ) = 10(x A x B ) = 10 9 = 90 (ii) Again begin with the point-slope form of the equation of a line: y y 1 = m(x x 1 ) Given the point P : (1, ), we have y = m(x 1) y = mx m y = mx + ( m) Since the line intersects the parabola y = 4 x at points A and B, set the line and parabola equal to each other: 4 x = mx + ( m) 0 = x + mx (m + ) Using the quadratic formula to solve for x yields x = m ± m (4)(1)[ (m + )] ()(1) Hence, = m ± m + 4m + 8 x A = m m + 4m + 8, and x B = m + m + 4m + 8 Therefore x A x B = m + 4m + 8 Substituting x A and x B into y = mx + ( m) yields the corresponding y-coordinates of A and B: y A = mx A + ( m) and y B = mx B + ( m) Therefore y A y B = m x A x B = m m + 4m + 8, and
4 4 MATH 41 FALL 009 HOMEWORK 3 SOLUTIONS AB = (x A x B ) + (y A y B ) = x A x B + y A y B = (xa x B ) + m (x A x B ) = (1 + m )(m + 4m + 8) Note that substituting m = 3 we obtain the results of part (i) Our results show that it is easy to minimize x A x B Indeed, completing the square in the quadratic polynomial gives: x A x B = m + 4m + 8 = (m + ) + 4 Hence, the smallest value of x A x B is 4 =, and it is attained when m = At this time of the course we do not have a good way of minimizing AB (iii) Again begin with the point-slope form of the equation of a line: Given P : (a, b), we have y y 1 = m(x x 1 ) y b = m(x a) y b = mx am y = mx + (b am) Since the line intersects the parabola y = 4 x at points A and B, in order to find the x-coordinates of the points of intersection, we solve the equation 4 x = mx + (b am) 0 = x + mx + (b am 4) Using the quadratic formula to solve for x yields x = m ± m (4)(1)(b am 4) ()(1) Hence, = m ± m + 4am + 4(4 b) x A = m m + 4am + 4(4 b), and x B = m + m + 4am + 4(4 b) Therefore x A x B = m + 4am + 4(4 b) Substituting x A and x B into y = mx+(b am) yields the corresponding y-coordinates of A and B: y A = mx A (b am) and y B = mx B (b am) Therefore y A y B = m x A x B = m m + 4am + 4(4 b), and
5 MATH 41 FALL 009 HOMEWORK 3 SOLUTIONS 5 AB = (x A x B ) + (y A y B ) = x A x B + y A y B = (xa x B ) + m (x A x B ) = (1 + m )(m + 4am + 4(4 b)) Our results show that it is easy to minimize x A x B Indeed, completing the square in the quadratic polynomial gives: x A x B = m + 4am + 4(4 b) = (m + a) + 4(4 b a ) Hence, the smallest value of x A x B is 4(4 b a ) = 4 b a, and it is attained when m = a Note that as P : (a, b) is in the interior of the parabola y = 4 x, b < 4 a Therefore 4 b a > 0, and the square root above exists Substituting a = 1, b =, we obtain the results of part (ii) At this time of the course we do not have a good way of minimizing AB Note that if P : (a, b) were on, or in the exterior of the parabola y = 4 x, then b 4 a Therefore 4 b a 0 In this case the tangent line(s) to the parabola through P would cut the chord AB of length 0 H3P3 (i) x 4 x (x )(x + x 6) x 4 x (x )(x + x 6) = (x + )(x ) x (x )(x + 3)(x ) = x + x (x )(x + 3) = (ii) x 4 x 3 + (x )(x + x 6) x 4 x 3 + (x )(x + x 6) = (x + )(x ) x 3 + (x )(x + 3)(x ) = x + x 3 + (x )(x + 3) = 5 6 (iii) x 3 x 3 x x 3 x 8x + 1 x x 3 x 8x + 1 = (x + 3)(x 3x + 9) x 3 (x + 3)(x )(x ) = x 3x + 9 x 3 (x ) = 7 5
6 6 MATH 41 FALL 009 HOMEWORK 3 SOLUTIONS (iv) x x 1 x x x 1 x = (x + 1)(x 6 x 5 + x 4 x 3 + x x + 1) x 1 (x + 1)(x 4 x 3 + x x + 1) x 6 x 5 + x 4 x 3 + x x + 1 = x 1 x 4 x 3 + x x + 1 = 7 5
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