Can anyone put this parabola into this standard form: x-h = a(y-k)^2
x^2-12x-48y-372=0 ?
Please help!
Sorry: Precalculus! :P
check your previous post of this problem
Thank you again! ^o^
To put the given equation of the parabola into the standard form x−h=a(y−k)², we need to complete the square.
The given equation is x² - 12x - 48y - 372 = 0.
First, let's rearrange the equation to group the x terms together and isolate the y term:
x² - 12x = 48y + 372.
Next, we want to complete the square for the x terms. To do this, take half the coefficient of x (which is -12 in this case), square it, and add it to both sides of the equation:
x² - 12x + (-12/2)² = 48y + 372 + (-12/2)².
Simplifying this, we get:
x² - 12x + 36 = 48y + 372 + 36.
The left side is now a perfect square:
(x - 6)² = 48y + 408.
Now, let's isolate the y term and divide both sides by 48:
(x - 6)²/48 = (48y + 408)/48.
Simplifying further:
(x - 6)²/48 = y + 17.
Finally, we can rewrite the equation in the standard form:
x - 6 = 48(y + 17).
Therefore, the parabola x² - 12x - 48y - 372 = 0 can be expressed in the standard form x - 6 = 48(y + 17)^2.