# SOLUTION: For the conic section described by the equation 9x^2+4y^2-18x+16y-11=0 What is the center? What is the length of the semimajor axes? What is the focal length?

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 Click here to see ALL problems on Quadratic-relations-and-conic-sections Question 616210: For the conic section described by the equation 9x^2+4y^2-18x+16y-11=0 What is the center? What is the length of the semimajor axes? What is the focal length?Answer by lwsshak3(7679)   (Show Source): You can put this solution on YOUR website!For the conic section described by the equation 9x^2+4y^2-18x+16y-11=0 What is the center? What is the length of the semimajor axes? What is the focal length? ** 9x^2+4y^2-18x+16y-11=0 complete the square 9x^2-18x+4y^2+16y-11=0 9(x^2-2x+1)+4(y^2+4y+4)=11+9+16 9(x-1)^2+4(y+2)^2=36 (x-1)^2/4+(y+2)^2/9=1 This is an equation for an ellipse with vertical major axis. Its standard form: ((x-h)^2/b^2+(y-k)^2/a^2=1, (a>b), (h,k)=(x,y) coordinates of the center For given equation: center: (1,-2) b^2=4 b=2 length of semi major or minor axis=2b=4 .. a^2=9 c^2=a^2-b^2=9-4=5 c=√5 focal length=22c=2√5