Answer :
To find the focus of a parabola given by the equation [tex]\( y = \frac{1}{8}(x^2 - 4x - 12) \)[/tex], we will first convert this equation to its vertex form, which is [tex]\( y = a(x - h)^2 + k \)[/tex].
1. Rewrite the equation in standard form:
Given equation:
[tex]\[ y = \frac{1}{8}(x^2 - 4x - 12) \][/tex]
2. Complete the square on the quadratic expression inside the parentheses:
- Start with the quadratic part [tex]\( x^2 - 4x \)[/tex].
- To complete the square, we add and subtract the same value. The required term to complete the square is [tex]\(\left( \frac{-4}{2}\right)^2 = 4\)[/tex].
Thus, rewrite [tex]\( x^2 - 4x \)[/tex]:
[tex]\[ x^2 - 4x = (x^2 - 4x + 4 - 4) = (x - 2)^2 - 4 \][/tex]
3. Substitute this completed square back into the equation:
[tex]\[ y = \frac{1}{8}[(x - 2)^2 - 4 - 12] = \frac{1}{8}(x - 2)^2 - \frac{1}{8} \cdot 16 \][/tex]
4. Simplify the equation:
[tex]\[ y = \frac{1}{8}(x - 2)^2 - 2 \][/tex]
Now the equation is in vertex form [tex]\( y = \frac{1}{8}(x - 2)^2 - 2 \)[/tex], where [tex]\( h = 2 \)[/tex] and [tex]\( k = -2 \)[/tex]. So, the vertex (h, k) of the parabola is [tex]\( (2, -2) \)[/tex].
5. Determine the value of [tex]\( a \)[/tex] and find the focal length:
For a parabola in the form [tex]\( y = a(x - h)^2 + k \)[/tex], the focal length (p) is given by [tex]\( \frac{1}{4a} \)[/tex]. Here, [tex]\( a = \frac{1}{8} \)[/tex].
[tex]\[ p = \frac{1}{4 \left( \frac{1}{8} \right) } = \frac{1}{\frac{1}{2}} = 2 \][/tex]
6. Find the coordinates of the focus:
For a parabola that opens upwards or downwards, the focus is located at [tex]\( (h, k + p) \)[/tex]. With [tex]\( h = 2 \)[/tex], [tex]\( k = -2 \)[/tex], and [tex]\( p = 2 \)[/tex]:
[tex]\[ \text{Focus} = (h, k + p) = (2, -2 + 2) = (2, 0) \][/tex]
So the correct answer is:
[tex]\[ \boxed{(2, 0)} \][/tex]
Therefore, the correct choice is:
D. [tex]\( (2, 0) \)[/tex]
1. Rewrite the equation in standard form:
Given equation:
[tex]\[ y = \frac{1}{8}(x^2 - 4x - 12) \][/tex]
2. Complete the square on the quadratic expression inside the parentheses:
- Start with the quadratic part [tex]\( x^2 - 4x \)[/tex].
- To complete the square, we add and subtract the same value. The required term to complete the square is [tex]\(\left( \frac{-4}{2}\right)^2 = 4\)[/tex].
Thus, rewrite [tex]\( x^2 - 4x \)[/tex]:
[tex]\[ x^2 - 4x = (x^2 - 4x + 4 - 4) = (x - 2)^2 - 4 \][/tex]
3. Substitute this completed square back into the equation:
[tex]\[ y = \frac{1}{8}[(x - 2)^2 - 4 - 12] = \frac{1}{8}(x - 2)^2 - \frac{1}{8} \cdot 16 \][/tex]
4. Simplify the equation:
[tex]\[ y = \frac{1}{8}(x - 2)^2 - 2 \][/tex]
Now the equation is in vertex form [tex]\( y = \frac{1}{8}(x - 2)^2 - 2 \)[/tex], where [tex]\( h = 2 \)[/tex] and [tex]\( k = -2 \)[/tex]. So, the vertex (h, k) of the parabola is [tex]\( (2, -2) \)[/tex].
5. Determine the value of [tex]\( a \)[/tex] and find the focal length:
For a parabola in the form [tex]\( y = a(x - h)^2 + k \)[/tex], the focal length (p) is given by [tex]\( \frac{1}{4a} \)[/tex]. Here, [tex]\( a = \frac{1}{8} \)[/tex].
[tex]\[ p = \frac{1}{4 \left( \frac{1}{8} \right) } = \frac{1}{\frac{1}{2}} = 2 \][/tex]
6. Find the coordinates of the focus:
For a parabola that opens upwards or downwards, the focus is located at [tex]\( (h, k + p) \)[/tex]. With [tex]\( h = 2 \)[/tex], [tex]\( k = -2 \)[/tex], and [tex]\( p = 2 \)[/tex]:
[tex]\[ \text{Focus} = (h, k + p) = (2, -2 + 2) = (2, 0) \][/tex]
So the correct answer is:
[tex]\[ \boxed{(2, 0)} \][/tex]
Therefore, the correct choice is:
D. [tex]\( (2, 0) \)[/tex]
