How do you write the standard form of the hyperbola −2x2+3y2+4x−60y+268=0?

2 Answers
Narad T.
Oct 17, 2016

−(x−1)2(√15)2+(y−10)2(√10)2=1

Explanation:

−2x2+3y2+4x−60y+268=0
Rearranging
−2(x2−2x)+3(y2−20y)=−268
Completing the squares
−2(x2−2x+1)+3(y2−20y+100)=−268−2+300
−2(x−1)2+3(y−10)2=30
Dividing by 30
−(x−1)215+(y−10)210=1
In standard form
−(x−1)2(√15)2+(y−10)2(√10)2=1

Douglas K.
Oct 17, 2016

Please see the explanation for the process.

(y−10)2(√10)2−(x−1)2(√15)2=1

Explanation:

Add 3k2−2h2−268 to both sides:

−2x2+4x−2h2+3y2−60y+3k2=3k2−2h2−268

Remove a factor of -2 from the first 3 terms and a factor of 3 from the next 3 terms:

−2(x2−2x+h2)+3(y2−20y+k2)=3k2−2h2−268

Using the pattern, (x−h)2=x2−2hx+h2, set the right side of the pattern equal to the first three terms:

x2−2hx+h2=x2−2x+h2

Combine like terms:

−2hx=−2x

Solve for h and compute h2:

h=1,h2=1

Substitute (x−1)2 for x2−2x+h2 and −2 for −2h2

−2(x−1)2+3(y2−20y+k2)=3k2−2−268

Using the pattern, (y−k)2=y2−2ky+k2, set the right side of the pattern equal to the second three terms:

y2−2ky+k2=y2−20y+k2

Combine like terms:

−2ky=−20y

k=10,k2=100

Substitute (y−10)2 for y2−20y+k2 and 300 for 3k2

−2(x−1)2+3(y−10)2=300−2−268

Combine the terms on the right:

−2(x−1)2+3(y−10)2=30

Divide both sides by 30:

(y−10)210−(x−1)215=1

Write the denominators as squares:

(y−10)2(√10)2−(x−1)2(√15)2=1

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