9th Edition

ISBN: 9780321716835

Author: Michael Sullivan

Publisher: Addison Wesley

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Question

Chapter 10, Problem 17RE

To determine

Tofind:the equation and if the equation is parabolic give its vertex, focus and directrix; if its ellipse gives its center, vertices and foci; if it’s is a hyperbola, give its center, vertices, foci and asymptotes.

Expert Solution & Answer

Answer to Problem 17RE

Thevertex is $(h,k)=(2,-1)$ , Focus= $(h,k-a)=(2,−1-1)=(2,−2)$ and Directrix= $y=k+a⇒y=-1+1=0$ .

Explanation of Solution

Given:

$4x2-16x+16y+32=0$ .

Concept used:

The standard form is $(x-h)2=4p(y-k)$ where the focus is $(h,k+p)$ and the directrix is $y=k-p$ . If the parabola is rotated so that its vertex is $(h,k)$ and its axis of symmetry is parallel to the x-axis, it has an equation of $(y-k)2=4p(x-h)$ where the focus is $(h+p,k)$ and the directrix is $x=h-p$ .

Calculation:

$4x2-16x+16y+32=0$ .

$4x2-16x+16+16y+16=0$ .

$4(x2-4x+4)+16(y+1)=0$ .

$(x2-4x+4)=-4(y+1)$ .

$(x-2)2=-4(y+1)$ .

Which is the equation of the parabola of the form:

$(x-h)2=-4a(y-k)$ .

$h=2, k=-1 and a=1$

The vertex is $(h,k)=(2,-1).$ .

Focus= $(h,k-a)=(2,−1-1)=(2,−2)$ .

Directrix= $y=k+a⇒y=-1+1=0$ .

$y=0$ .

Hence, vertex is $(h,k)=(2,-1)$ , Focus= $(h,k-a)=(2,−1-1)=(2,−2)$ and Directrix= $y=k+a⇒y=-1+1=0$ .

Chapter 10, Problem 16RE

Chapter 10, Problem 18RE

Chapter 10 Solutions

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Tofind: the equation and if the equation is parabolic give its vertex, focus and directrix; if its ellipse gives its center, vertices and foci; if it’s is a hyperbola, give its center, vertices, foci and asymptotes.