Mathematics, 17.07.2019 20:00 tynitenaire

What are the coordinates of the focus of the parabola? y=−1/16x^2 − x + 2 (−8, 2) (8, 2) (8, 10) (−8, 10)

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This is a tough one. the general form of a parabola is $(x-h) ^{2} =4p(y-k)$ , 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: $- \frac{1}{16} x^{2} -x=-2$ . In order to complete the square, the leading coefficient on the squared term has to be a +1. Ours is a $- \frac{1}{16}$ , so we have to factor that out of the x terms. When you do that you end up with $- \frac{1}{16} ( x^{2} +16x)=-2$ . 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 $- \frac{1}{16}$ is still hanging out in front, which means that when we add in 64, we are actually adding in $- \frac{1}{16} *64$ which is -4. Now here's what we have: $- \frac{1}{16} ( x^{2} +16x+64)=-2-4$ which simplifies to $- \frac{1}{16}( x^{2} +16x+64)=-6$ . 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 $- \frac{1}{16} (x+8) ^{2} =-6$ . Now just for simplicity, we will take baby steps. Move the -6 back over by addition and set it back equal to y: $- \frac{1}{16}(x+8) ^{2}+6=y$ . Now we will work on getting into standard form. Move the 6 back over by the y (baby steps, remember) to get $- \frac{1}{16} (x+8) ^{2} =y-6$ . Multiply both sides by -16 to get our "p" on the right: $(x+8) ^{2} =-16(y-6)$ . 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!

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2 x t divided by n

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536

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What are the coordinates of the focus of the parabola? y=−1/16x^2 − x + 2 (−8, 2) (8, 2) (8, 10) (−...

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