4x^{2}+25y^{2}=1^{
}
how to solve an ellipse
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1 Answer
Write in standard form as x^{2}/(1/4)+y^{2}/(1/25)=1
The major axis is of length 2√1/4=2×1/2=1
The minor axis is of length 2√1/25=2×1/5=2/5
The center is at (0,0)
The focal length is determined by c=√(1/4-1/25)=√((25-4)/100)=(√21)/10
It crosses the x-axis at x^{2}=1/4 or at x=±1/2
The foci are at x=±(√21)/10,y=0
The eccentricity is c/a=((√21)/10)/(1/2)=(√21)/5
What else do you mean by solve?