Determine the vertices, foci and eccentricity of the ellipse 5x2+6y2=30. Determine the lengths of the major and minor
axes, and sketch the graph.
If we divide both sides by 16, then we have
x26+y25=1
We'll see that the function has the form x2a2+y2b2=1. Since, the denominator of x2 is
larger, then the ellipse has a horizontal major axis. This gives a2=6 and b2=5, so c2=a2−b2=6−5=1. Thus, a=√6, b=√5
and c=1. Then the following is determined as.
Vertices(±a,0)→(±√6,0)Foci(±c,0)→(±1,0)Eccentricity (e)ca→1√6Length of major axis2a→2√6Length of minor axis2b→2√5
Monday, April 2, 2018
College Algebra, Chapter 8, 8.2, Section 8.2, Problem 16
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