Find the Properties (x^2)/16+(y^2)/25=1
Problem
Solution
Identify the type of conic section. Since the equation is in the form
(x2)/(b2)+(y2)/(a2)=1 witha2>b2 this is a vertical ellipse centered at the origin(0,0) Determine the values of
a andb We havea2=25 soa=5 We haveb2=16 sob=4 Calculate the distance to the foci
c using the relationshipc2=a2−b2
Identify the vertices and co-vertices. The vertices are located at
(0,±a) which are(0,5) and(0,−5) The co-vertices are located at(±b,0) which are(4,0) and(−4,0) Identify the foci. The foci are located at
(0,±c) which are(0,3) and(0,−3) Calculate the eccentricity
e using the formulae=c/a
Final Answer
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