Question

for the function $f(x) = -\frac{1}{8}(x + 2)^2 + 3$, what is the focal point? what is the equation of the directrix?
focal point: ( select , select )
equation of the directrix: select = select

Explanation:

Step1: Recall the vertex form of a parabola

The vertex form of a parabola that opens up or down is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. For our function \( f(x)=-\frac{1}{8}(x + 2)^2+3 \), we can rewrite it as \( y=-\frac{1}{8}(x - (-2))^2+3 \), so the vertex \((h,k)=(-2,3)\).

Step2: Find the value of \( p \)

For a parabola in the form \( y = a(x - h)^2 + k \), the relationship between \( a \) and \( p \) (the distance from the vertex to the focus and from the vertex to the directrix) is \( a=\frac{1}{4p} \). Solving for \( p \), we get \( p=\frac{1}{4a} \).

In our function, \( a = -\frac{1}{8} \). Substitute \( a \) into the formula for \( p \):

\( p=\frac{1}{4\times(-\frac{1}{8})}=\frac{1}{-\frac{1}{2}}=- 2 \)? Wait, no, wait. Wait, the formula is \( a=\frac{1}{4p} \), so \( p=\frac{1}{4a} \). Let's compute that again. \( a = -\frac{1}{8} \), so \( 4a=4\times(-\frac{1}{8})=-\frac{1}{2} \), then \( p=\frac{1}{4a}=\frac{1}{-\frac{1}{2}}=-2 \)? Wait, no, that can't be. Wait, actually, the sign of \( a \) tells us the direction. If \( a>0 \), the parabola opens up; if \( a < 0 \), it opens down. The distance from the vertex to the focus is \( |p| \), and \( p=\frac{1}{4a} \). Wait, let's do it correctly.

Given \( y = a(x - h)^2 + k \), the focus is at \((h,k + p)\) and the directrix is \( y=k - p \), where \( p=\frac{1}{4a} \).

So \( a = -\frac{1}{8} \), so \( p=\frac{1}{4\times(-\frac{1}{8})}=\frac{1}{-\frac{1}{2}}=-2 \). Wait, but \( p \) is the distance, but since \( a \) is negative, the parabola opens down, so the focus is below the vertex (since it opens down). Wait, no: if \( a<0 \), the parabola opens downward, so the focus is \( |p| \) units below the vertex, and the directrix is \( |p| \) units above the vertex. Wait, let's re - express the formula.

Actually, the correct formula is: for \( y=a(x - h)^2 + k \), the focus is \((h,k+\frac{1}{4a})\) and the directrix is \( y = k-\frac{1}{4a} \). Let's use this.

So \( \frac{1}{4a}=\frac{1}{4\times(-\frac{1}{8})}=\frac{1}{-\frac{1}{2}}=-2 \). Wait, but distance can't be negative in terms of direction. Wait, the sign of \( \frac{1}{4a} \) tells us the direction. Since \( a = -\frac{1}{8}<0 \), the parabola opens downward, so the focus is below the vertex. So the \( y \) - coordinate of the focus is \( k+\frac{1}{4a}=3+(-2)=1 \). The \( x \) - coordinate of the focus is the same as the \( x \) - coordinate of the vertex, which is \( h=-2 \). So the focus (focal point) is \((-2,1)\).

For the directrix, since the directrix is a horizontal line (because the parabola opens up or down) and is \( |\frac{1}{4a}| \) units away from the vertex in the opposite direction of the focus. The equation of the directrix is \( y=k-\frac{1}{4a} \)? Wait, no. Wait, if the focus is at \( (h,k + p) \) and the directrix is \( y=k - p \), where \( p=\frac{1}{4a} \). Wait, let's take an example. If \( a=\frac{1}{4} \), then \( p = 1 \), focus is \( (h,k + 1) \), directrix \( y=k - 1 \). If \( a=-\frac{1}{4} \), then \( p=-1 \), focus is \( (h,k-1) \), directrix \( y = k + 1 \). Ah, I see my mistake earlier. The correct formula is \( p=\frac{1}{4a} \), so when \( a \) is negative, \( p \) is negative, meaning the focus is below the vertex (for vertical parabola opening down) and the directrix is above the vertex.

So let's recalculate \( p \) correctly. \( a=-\frac{1}{8} \), so \( p=\frac{1}{4a}=\frac{1}{4\times(-\frac{1}{8})}=\frac{1}{-\frac{1}{2}}=-2 \). So the focus (focal point) is at \( (h,k + p)=(-2,3+(-2))=(-2,1) \).

The directrix is a horizontal line. The distance from the vertex to the directrix is \( |p| \), and since the parabola opens downward, the directrix is above the vertex. The equation of the directrix is \( y=k - p \)? Wait, no. Wait, if \( p=\frac{1}{4a} \), and for a parabola \( y = a(x - h)^2 + k \), the directrix is \( y=k-\frac{1}{4a} \) when \( a>0 \)? No, let's use the standard form. The standard form of a vertical parabola is \( (x - h)^2 = 4p(y - k) \). Let's rewrite our equation in standard form.

Starting from \( y=-\frac{1}{8}(x + 2)^2+3 \), subtract 3 from both sides: \( y - 3=-\frac{1}{8}(x + 2)^2 \). Multiply both sides by - 8: \( - 8(y - 3)=(x + 2)^2 \), or \( (x + 2)^2=-8(y - 3) \). Now, the standard form \( (x - h)^2 = 4p(y - k) \), so comparing \( (x + 2)^2=-8(y - 3) \) with \( (x - h)^2 = 4p(y - k) \), we have \( 4p=-8 \), so \( p=-2 \). Ah, this is a better way. So \( 4p=-8\implies p = - 2 \).

So the vertex is \((h,k)=(-2,3)\). The focus of a parabola in the form \( (x - h)^2 = 4p(y - k) \) is at \( (h,k + p) \). So substituting \( h=-2 \), \( k = 3 \), and \( p=-2 \), we get the focus (focal point) as \( (-2,3+(-2))=(-2,1) \).

The directrix of a parabola in the form \( (x - h)^2 = 4p(y - k) \) is the line \( y=k - p \). Substituting \( k = 3 \) and \( p=-2 \), we get \( y=3-(-2)=3 + 2=5 \). So the equation of the directrix is \( y = 5 \).

Answer:

Focal Point: \((-2, 1)\)
Equation of the Directrix: \( y = 5 \)