What is the standard form of this parabola:
-2x^2+16x+24y-224=0
Please Help, parabolas are evil.
Divide both sides by -2 and complete the square for the x^2 binomial.
x^2 -8x -12y +112 = 0
x^2 -8x +16 -12y +96 = 0
(x -4)^2 -12(y-8) = 0
solution:you must be solved for the y of way implicit
-2x^2+16x-224=-24y
2x^2-16x+224=24y
(1/12)x^2-(2/3)x+(224/24)=y
and solve it
To find the standard form of a parabola, you need to complete the square. The standard form of a parabola is written as follows:
y = a(x-h)^2 + k
Where (h, k) represents the vertex of the parabola and "a" is a constant that determines whether the parabola opens upwards (if a > 0) or downwards (if a < 0).
Let's begin by isolating the terms with "x" and "y" on one side of the equation:
-2x^2 + 16x + 24y - 224 = 0
Next, rearrange the equation so that the terms involving "x" and "y" are together:
-2x^2 + 16x + 24y = 224
Now, divide the entire equation by the coefficient of x^2 (which is -2) to create a leading coefficient of 1:
x^2 - 8x - 12y = -112
To complete the square, focus on the terms involving "x." Take half of the coefficient of x (-8) and square it:
(-8/2)^2 = 16
Add this value (16) to both sides of the equation:
x^2 - 8x + 16 - 12y = -112 + 16
By adding 16 to the left side of the equation, we've made it possible to factor the left side as a perfect square:
(x - 4)^2 - 12y = -96
Now, rearrange the equation to match the standard form:
(x - 4)^2 = 12y - 96
Finally, divide both sides of the equation by the constant term (which is 12) to isolate y:
12y - 96 = (1/12)(x - 4)^2
12y = (1/12)(x - 4)^2 + 96
Simplify further to obtain the standard form:
y = (1/12)(x - 4)^2 + 8
So, the standard form of the given parabola is y = (1/12)(x - 4)^2 + 8.