Explanation Formula used: Parabola with vertex at ( h , k ) ; Axis of symmetry parallel to a coordinate axis a > 0 . Vertex Focus Directrix Equation Axis of symmetry Opens ( h , k ) ( h + a , k ) x = h − a ( y − k ) 2 = 4 a ( x − h ) Parallel to x -axis Right ( h , k ) ( h − a , k ) x = h + a ( y − k ) 2 = − 4 a ( x − h ) Parallel to x -axis Left ( h , k ) ( h , k + a ) y = k − a ( x − h ) 2 = 4 a ( y − k ) Parallel to y -axis Up ( h , k ) ( h , k − a ) y = k + a ( x − h ) 2 = − 4 a ( y − k ) Parallel to y -axis Down Calculation: The equation of the parabola is, x 2 − 4 x + 10 y + 4 = 0 . To obtain the standard form of the parabola, separate the terms involving x and y as x 2 − 4 x = − 10 y − 4 . Now, apply the completing the square method and obtain the standard form of the parabola. x 2 − 4 x = − 10 y − 4 ( x 2 − 4 x + 2 2 − 2 2 ) = − 10 y − 4 ( x − 2 ) 2 = − 10 y − 4 + 4 ( x − 2 ) 2 = − 10 y Thus, the equation of the parabola is ( x − 2 ) 2 = − 10 y . Note that the equation is of the form ( x − h ) 2 = − 4 a ( y − k ) which opens down with vertex at ( h , k ) , focus at ( h , k − a ) , and directrix y = k + a

4th Edition

ISBN: 9780134556079

Author: Sullivan, Michael, III, Struve, Katherine R., Mazzarella, Janet.

Publisher: Pearson,

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Have you ever seen a bridge in the shape of a parabola? Where is its maximum on the bridge? The maximum of that bridge is the highest point on that bridge. The process of finding the maximum of the bridge here is called the maximization.

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Chapter 12.3, Problem 50E

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The vertex, focus, directrix of the parabola x2−4x+10y+4=0 and graph the same.

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Chapter 12.3, Problem 49E

Chapter 12.3, Problem 51E

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The vertex, focus, directrix of the parabola x 2 − 4 x + 10 y + 4 = 0 and graph the same.

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