Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 11 - Section 11.4 - Shifted Conics - 11.4 Exercises - Page 814: 17

Answer

(a) Vertex: $V(1,0)$ Focus: $F(1,\frac{1}{8})$ Directrix: $y=-\frac{1}{8}$ (b)

Work Step by Step

Equation of a parabola with vertical axis and vertex at $(h,k)$: $(x-h)^2=4p(y-k)$ $2(x-1)^2=y$ $(x-1)^2=\frac{1}{2}(y-0)$ $h=1$ $k=0$ Vertex: $V(1,0)$ $4p=\frac{1}{2}$ $p=\frac{1}{8}$. Parabola opens upward. The given equation can be obtained by shifting $x^2=\frac{1}{2}y$ right 1 unit. In this equation: Focus: $F(0,p)=F(0,\frac{1}{8})$ Directrix: $y=-p=-\frac{1}{8}$ Now, shift the vertex 1 unit to the right: Focus: $F(1,\frac{1}{8})$ Directrix: $y=-\frac{1}{8}$
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