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Q55.

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Precalculus Mathematics for Calculus
Found in: Page 789
Precalculus Mathematics for Calculus

Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508

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Short Answer

Finding the Equation of a Parabola

Find an equation of the parabola whose graph is shown

The equation of parabola is y2=x

See the step by step solution

Step by Step Solution

Step 1. Given information

Graph of parabola is given as-

Step 2. Concept used

The equation of a parabola having a vertex at origin is –

y2=4px

Where, p>0 then parabola opens to the right

p<0 then parabola opens to the left.

The directrix is the right side of y-axis is so, which means the focus should be negative, the parabola opens to the left and the coordinates of focus are obtained. On substituting the value of p in the equation y2=4px to obtain an equation of a parabola.

Step 3. Calculation

Substituting the point (4,-2) in the equation of parabola y2=4px, we get –

(-2)2=4p(4)

4=16p16p=4p=416p=14

Now, substituting the value of p in equation y2=4px , we get the equation as.

y2=414xy2=x

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