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    Y X BX C: Analytical Geometry: Solutions

    Uploaded by

    Marcus Pranga
    This document provides 11 problems involving analytical geometry concepts such as parabolas, circles, ellipses, hyperbolas, and their equations. The problems ask the reader to determine values like the vertex of a parabola, the distance between circle centers, which quadrants a circle passes through, the coefficient of a quadratic polynomial, and the center and vertices of an ellipse. By solving the systems of equations for each problem, values like b-c, the distance between circles, and the coefficient or constant term in various equations can be determined.

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    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd

    Y X BX C: Analytical Geometry: Solutions

    Uploaded by

    Marcus Pranga
    44 views2 pages
    This document provides 11 problems involving analytical geometry concepts such as parabolas, circles, ellipses, hyperbolas, and their equations. The problems ask the reader to determine values like the vertex of a parabola, the distance between circle centers, which quadrants a circle passes through, the coefficient of a quadratic polynomial, and the center and vertices of an ellipse. By solving the systems of equations for each problem, values like b-c, the distance between circles, and the coefficient or constant term in various equations can be determined.

    Original Description:

    Original Title

    AnalyticalGeometry_S

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    This document provides 11 problems involving analytical geometry concepts such as parabolas, circles, ellipses, hyperbolas, and their equations. The problems ask the reader to determine values like the vertex of a parabola, the distance between circle centers, which quadrants a circle passes through, the coefficient of a quadratic polynomial, and the center and vertices of an ellipse. By solving the systems of equations for each problem, values like b-c, the distance between circles, and the coefficient or constant term in various equations can be determined.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    44 views2 pages

    Y X BX C: Analytical Geometry: Solutions

    Uploaded by

    Marcus Pranga
    This document provides 11 problems involving analytical geometry concepts such as parabolas, circles, ellipses, hyperbolas, and their equations. The problems ask the reader to determine values like the vertex of a parabola, the distance between circle centers, which quadrants a circle passes through, the coefficient of a quadratic polynomial, and the center and vertices of an ellipse. By solving the systems of equations for each problem, values like b-c, the distance between circles, and the coefficient or constant term in various equations can be determined.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 2

    Analytical Geometry:

    Solutions:
    1. The vertex of the graph of the parabola y = 3 x 2 + bx + c is at (-2,-7); determine
    the value of b-c; NCALGII2002:33
    2. The distance between the centers of the circles x 2 − 8 x + y 2 − 28 y + 112 = 0 and
    x 6 + 6 x + y 2 + 20 y − 12 = 0 is: FUR2003JR11
    3. Through which quadrants does the circle x 2 + 4 x + y 2 − 6 y + 1 = 0 pass?
    FUR2003JR13
    4. If a parabola has vertex (0,0) and focus (1,0), the equation is: FUR2003JR!5
    5. If Q is the point on the circle x 2 − 10 x + y 2 + 6 y + 29 = 0 which is furthest from
    the point P(−1, −6) , then the distance from P to Q is: FUR2003SR2.

    x 2 − 10 x + y 2 + 6 y + 29 = 0 ⇔ x 2 − 10 x + 25 + y 2 + 6 y + 9 = −29 + 25 + 9 = 5
    ⇔ ( x − 5 ) + ( y + 3) = 5
    2 2

    So this is a circle with center (5,-3) and radius 5 . The distance from P to Q
    includes the distance from P to the center and then an additional radius. So this
    ( 5 − ( −1) ) + ( −3 − ( −6 ) )
    2 2
    distance is + 5 = 62 + 32 + 5 = 45 + 5

    6. The parabola y = ax 2 + bx + c intersects the y-axis in the point (0,8) and intersects
    the x-axis in the single point (2,0). How many pairs of integers (m,n) with
    −2000 ≤ m ≤ 2000 lie on the parabola? NCSMC2000INT4

    The parabola must be of the form


    y = a ( x − 2 ) ⇒ 8 = a ( 0 − 2 ) ⇒ a = 2 ⇒ y = 2 ( x − 2 ) . Since the y-values grow
    2 2 2

    much more quickly than the x-values, we need to find all possible points whose y-
    values are less than or equal to 2000. So
    2 ( x − 2 ) ≤ 2000 ⇒ ( x − 2 ) ≤ 1000 ⇒ − 1000 ≤ x − 2 ≤ 1000 . But since we
    2 2

    want integer coordinates, this means that −31 ≤ x − 2 ≤ 31 ⇒ −29 ≤ x ≤ 33 , so


    there are 63 such lattice points.

    7. Let f(x) be a quadratic polynomial such that f(3) = 15 and f(-3) = -9. Find the
    coefficient of x in f(x). SMC2002MC1

    We know that 15 = a ( 32 ) + b(3) + c and −9 = a ( −3) + b(−3) + c , and since we are


    2

    looking for b, subtract these to get 24 = −6b ⇒ b = −4.

    8. The center C and the vertices (V) of the ellipse 4 x 2 + 9 y 2 − 16 x − 54 y + 61 = 0 are:


    NCALGII2000#9

    1
    Written & Compiled by John Goebel, NCSSM Problem Solving Course, 2006
    4 x 2 + 9 y 2 − 16 x − 54 y + 61 = 0 ⇔ 4 ( x 2 − 4 x + 4 ) + 9 ( y 2 − 6 y + 9 ) = −61 + 16 + 81 = 36

    ( x − 2) ( y − 3)
    2 2

    ⇔ 4 ( x − 2 ) + 9 ( y − 3) = 36 ⇔ +=1
    2 2

    9 4
    so the center of this ellipse is (2,3) and vertices (end points of the major axis) are
    (5,3) and (-1,3).

    9. The quadratic equation y = ax 2 + bx + c is known to pass through the points (0,5),


    (2,11) and (-2,15). Find the sum of a and b. NCALGII2000#17

    We know that 11 = 4a + 2b + c,5 = c,15 = 4a − 2b + c , so


    4a + 2b = 6, 4a − 2b = 10 ⇒ 8a = 16 ⇒ a = 2 , so b = −1 and the sum a + b = 1 .

    10. Write an equation for the hyperbola with horizontal transverse axis and
    asymptotes y = ± 4 x . NCALGII2000#25
    3

    x2 y2
    − =1
    9 16

    11. A parabola of the form y = x 2 + bx + c contains the points (2,3) and (4,3), Find
    the value of c. AMC102006.

    We know that 3 = 4 + 2b + c,3 = 16 + 4b + c , so


    2b + c = −1, 4b + c = −13 ⇒ 2b = −12, b = −6 . This makes c = 11 .

    2
    Written & Compiled by John Goebel, NCSSM Problem Solving Course, 2006

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