Hyperbola: Rewrite the equation in standard form

Question

2y² - 3x² - 16y - 12x -10 = 0

Answer 1

Hey! Lets solve this shall we ?

So, we must take what we are given and turn it into the standard form of a hyperbola, right! Beginning from

2y² - 3x² - 16y - 12x - 10=0

We must rearrange the equation to match up the like variables such that

2y² - 16y - 3x² - 12x = 10............Plus adding 10 to the right side, right ? Now, we must group together certain terms so can begin Completing the Square such that,

(2y² - 16y) + (-3x² - 12x) = 10.............Now we factor out certain numbers in each group such that

2(y² - 8y) - 3(x² + 4x) = 10...........See? Now we begin the Completing the Square process by taking half of

8 and 4 and squaring them both getting 16 and 4 such that

2(y² - 8y + 16) - 3(x² + 4x + 4) = 10 + 16 + 4............Notice that what we do on the inside of the parentheses

must be added to ten as well. Then, we factor the polynomials such that

2(y - 4)² - 3(x + 2)² = 30...........Finally, we divide 30 to both sides such that

(y-4)² / 15 - (x + 2)² / 10 = 1

I hope this helped!