Question

The general form of a hyperbola is 6x2−5y2+12x+50y−149=0.

What is the standard form of this hyperbola?

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Answer 1

The hyperbola 6x² - 5y² + 12x + 50y - 149 = 0 represented in standard form is (x + 1)²/5 - (y - 5)²/6 = 1

Converting the hyperbola equation to standard form

From the question, we have the following parameters that can be used in our computation:

6x² - 5y² + 12x + 50y - 149 = 0

Rewrite as

6x² - 5y² + 12x + 50y = 149

Group the equation by the variables

6x² + 12x - 5y² + 50y = 149

So, we have

6(x² + 2x) - 5(y² - 10y) = 149

Complete the square on x and y

So, we have

6(x + 1)² - 5(y - 5)² = 149 + 6 - 125

This gives

6(x + 1)² - 5(y - 5)² = 30

Next, we divide through by 30

(x + 1)²/5 - (y - 5)²/6 = 1

Hence, the standard form is (x + 1)²/5 - (y - 5)²/6 = 1