Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 11: Parametric Equations and Polar Coordinates - Section 11.6 - Conic Sections - Exercises 11.6 - Page 678: 47

Answer

$(x-1)^{2}=8(y+7)$ New focus: $\quad (1, -5)$ New directrix: $\quad y=-9$ New vertex:$ \quad (1,-7)$

Work Step by Step

Shift right by $1 \Rightarrow$ in the equation, replace $x$ with $x-1$ Shift down by $7 \Rightarrow$ in the equation, replace $y$ with $y+7$ New equation:$\quad (x-1)^{2}=8(y+7)$ $x^{2}=8x\quad $is of the form $x^{2}=4py $ ,$\qquad$ (opens up), $p=2$ focus: $\quad(0,p)$= $\quad(0,2)$ directrix: $ y=-p\Rightarrow \quad y=-2$ vertex:$\quad (0,0)$ The translations are such that $(x,y)\rightarrow(x',y')$, where $\left\{\begin{array}{ll} x'=x+1 & \text{... shift right}\\ y'=y-7 & \text{... shift down} \end{array}\right.$ New focus: $\quad(0+1,2-7) = (1, -5)$ New directrix: $ y=-2-7\Rightarrow \quad y=-9$ New vertex:$\quad (0+1,0-7) = (1,-7)$
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