This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ertices are (7, –3) and (7, –5). Since a 2 + b 2 = c2, then
, which tells us that the foci are
and
. The asymptotes are
the lines that pass through the center and have slopes equal to ± a/b. Those are
and
. Here is a sketch.
y
x 4. Find the equation of the line tangent to the ellipse through the point . We need a point and a slope. We have a point,
. To get the slope we should evaluate dy/dx at the
point in question. So, to get started, I need dy/dx, which I can get using implicit differentiation. So the slope of the tangent line is . And the (unsimplified) equation of the tangent line is . 5. The region bounded by the hyperbola
and the vertical line through its rightmost focus is
revolved about the xaxis. Set up the definite integral needed to find the volume of the resulting solid. (You
do not need to evaluate this integral.)
First a sketch. In preparation for the sketch:
From the formula I can "read off"
.
So the center is (0, 0), the vertices are at (–3, 0) and (3, 0), and the foci are at
and
And the hyperbola has asymptotes
, which you really don't need for the purposes...
View
Full
Document
This document was uploaded on 03/18/2014 for the course MATH 237 at Frostburg.
 Fall '1
 Staff
 Calculus

Click to edit the document details