Respuesta :
This is a tough one. the general form of a parabola is [tex](x-h) ^{2} =4p(y-k)[/tex], where h and k are the coordinates of the vertex and p is the distance from the vertex to the focus. In order to get our parabola into this form and solve for p (which will give us our focal point), we have to complete the square. Set the parabola equal to 0, then move over the constant to get this equation: [tex]- \frac{1}{16} x^{2} -x=-2[/tex]. In order to complete the square, the leading coefficient on the squared term has to be a +1. Ours is a [tex]- \frac{1}{16} [/tex], so we have to factor that out of the x terms. When you do that you end up with [tex]- \frac{1}{16} ( x^{2} +16x)=-2[/tex]. Now we can complete the square by taking half the linear term, squaring it, and adding it to both sides. Our linear term is 16, so half of 16 is 8 annd 8 squared is 64. HOWEVER, on the left side, that [tex]- \frac{1}{16} [/tex] is still hanging out in front, which means that when we add in 64, we are actually adding in [tex]- \frac{1}{16} *64[/tex] which is -4. Now here's what we have: [tex]- \frac{1}{16} ( x^{2} +16x+64)=-2-4[/tex]which simplifies to [tex]- \frac{1}{16}( x^{2} +16x+64)=-6 [/tex]. Creating the perfect square binomial on the left was the point of this (to give us our vertex), so when we do that we have [tex]- \frac{1}{16} (x+8) ^{2} =-6[/tex]. Now just for simplicity, we will take baby steps. Move the -6 back over by addition and set it back equal to y: [tex]- \frac{1}{16}(x+8) ^{2}+6=y [/tex]. Now we will work on getting into standard form. Move the 6 back over by the y (baby steps, remember) to get [tex]- \frac{1}{16} (x+8) ^{2} =y-6[/tex]. Multiply both sides by -16 to get our "p" on the right: [tex](x+8) ^{2} =-16(y-6)[/tex]. We need to use our "4p" part of the standard form to find the p, which is the distance from the vertex to the focus. 4p=-16, and p = -4. That means that the focus is 4 units below the vertex. Let's figure out what the vertex is. From our equation, the vertex is ( -8, 6), and since this is an upside-down opening parabola, the focus will be aligned with the x-coordinate of the vertex. So our focus lies 4 units below 6 (6 is the y coordinate of the vertex which indicates up and down movement), so our focus has coordinates of (-8, 2), the first choice above. Told you it was a tough one! These conics are quite challenging!
The correct option is Option A: the coordinate of the focus of the parabola is (-8,2).
What is focus of the parabola?
In parabola every point on the parabola is at a equal distance from the fixed point and fixed line. that fixed point is called focus.
How to find the focus of the parabola?
If the equation of the parabola is given by the equation y=a(x-h)²+k
where h, k is the center of the parabola and the coordinate of the focus will be (h, k+1/(4a))
here, the equation of parabola is given by
y=−1/16x² − x + 2
converting the x term in perfect square
⇒y=−(1/16x² + x )+2
⇒y=−((1/4x)² +(2*1/4*2 x)+2²-2² )+2
⇒y=−((1/4x)² +(2*1/4*2 x)+2²-4 )+2
⇒y=−((1/4x)² +(2*1/4*2 x)+2²)+4+2
⇒y=−((1/4x)² +(2*1/4*2 x)+2²)+6
⇒y=−(1/4x+2)²+6
⇒y=−(x+8)²/16+6
comparing with the Parabola equation y=a(x-h)²+k
a=-1/16
h=-8
k=6
the coordinate of the focus will be (h, k+1/(4a))
x coordinate of the focus = -8
y coordinate of the focus = k+1/(4a) = 6+1/(4(-1/16))=6-4=2
coordinate of the focus will be (-8,2).
Learn more about the focus of the parabola
here: https://brainly.com/question/11844921
#SPJ2
