College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 6 - Section 6.1 - Parabolas - 6.1 Exercises - Page 590: 28

Answer

Focus = $\left(-\dfrac{1}{64}, 0\right)$; Directrix $x=\dfrac{1}{64}$; Axis of Symmetry $y=0$.

Work Step by Step

Bring the given equation to the form $y^2=4px$: $y^2=-\frac{1}{16}x$ Here, $4p=-\dfrac{1}{16} \implies p=-\dfrac{1}{64}$ and vertex $(h,k)=(0,0)$ Since, the parabola is horizontal, the axis of symmetry is $y=0$ with focus $(p,0)=\left(-\dfrac{1}{64}, 0\right)$ Now, the directrix is $x=-p=-\left(-\dfrac{1}{64}\right)=\dfrac{1}{64}$ Our results are: Focus = $\left(-\dfrac{1}{64}, 0\right)$; Directrix $x=\dfrac{1}{64}$; Axis of Symmetry $y=0$.
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