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Identify the conic with equation 16y^2-x^2+x-4y-9+0CircleParabolaHyperbolaEllipseThis is an assgement I don’t understand I need an explanation please

Identify the conic with equation 16y^2-x^2+x-4y-9+0CircleParabolaHyperbolaEllipseThis-example-1
User Chucky
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1 Answer

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we have the equation


16y^2-x^2+x-4y-9=0

step 1

Group similar terms and move the constant term to the right side


(16y^2-4y)+(-x^2+x)=9

step 2

Factor 16 in the first term and factor -1 in the second term


16(y^2-(y)/(4))-(x^2-x)=9

step 2

Complete the square twice


16(y^2-(y)/(4)+(1)/(64)-(1)/(64))-(x^2-x+(1)/(4)-(1)/(4))=9
16(y^2-y\/4+1\/64)-(x^2-x+1\/4)=9+(1)/(4)-(1)/(4)

step 3

Rewrite as perfect squares


16(y-(1)/(8))^2-(x-(1)/(2))^2=9

step 4

Divide both sides by 9


(16(y-1\/8)^2)/(9)-((x-1\/2)^2)/(9)=1

therefore

The answer is Hyperbola

User Johanna Larsson
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