Question

the parabola has the equation $y^2 - 6y + 24x - 111 = 0$ find the distance between the focus and the origin power of 2

Explanation:

Step1: Rewrite the parabola equation

First, we complete the square for the \( y \)-terms in the equation \( y^{2}-6y + 24x-111 = 0 \).

For the \( y \)-terms: \( y^{2}-6y=(y - 3)^{2}-9 \) (using the formula \( (a - b)^2=a^{2}-2ab + b^{2} \), here \( a = y \), \( 2b=6\Rightarrow b = 3 \))

Substitute back into the equation:
\( (y - 3)^{2}-9+24x-111 = 0 \)
Simplify: \( (y - 3)^{2}+24x-120 = 0 \)
Rearrange to get the standard form of a parabola that opens horizontally: \( (y - 3)^{2}=-24x + 120 \)
\( (y - 3)^{2}=-24(x - 5) \)

Step2: Identify the vertex and focus

The standard form of a horizontal parabola is \( (y - k)^{2}=4p(x - h) \), where \((h,k)\) is the vertex and \(p\) is the distance from the vertex to the focus (and also from the vertex to the directrix).

Comparing \( (y - 3)^{2}=-24(x - 5) \) with \( (y - k)^{2}=4p(x - h) \), we have:
\( 4p=-24\Rightarrow p=- 6 \)
The vertex \((h,k)=(5,3)\)

For a horizontal parabola \( (y - k)^{2}=4p(x - h) \), if \( p<0 \), the parabola opens to the left. The focus is at \((h + p,k)\)

So the focus coordinates are: \( h + p=5+(-6)=-1 \), \( k = 3 \). So the focus is at \((-1,3)\)

Step3: Calculate the distance squared from focus to origin

The distance \( d \) between two points \((x_1,y_1)\) and \((x_2,y_2)\) is given by \( d=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}} \), and the distance squared is \( d^{2}=(x_2 - x_1)^{2}+(y_2 - y_1)^{2} \)

The origin is \((0,0)\) and the focus is \((-1,3)\)

So the distance squared \( d^{2}=(-1 - 0)^{2}+(3 - 0)^{2}=(-1)^{2}+3^{2}=1 + 9=10 \)

Answer:

\( 10 \)