Question

Scale: 1 unit = 10 meters

The hyperbolic cross section of a cooling tower is given by the equation 4x^2 − y^2 + 16y − 80 = 0. The center of the cooling tower is the same as the center of the hyperbola, and the x-axis represents the ground surface.

The diameter at the center of the tower is___meters. The center of the tower is___meters above the ground.

Answer 1

@Michele_Laino @triciaal

Answer 2

Here is the general formula for a traslated hyperbola:
\[A{x^2} + B{y^2} + Cx + Dy + E = 0\]

Answer 3

bycomparison with your formula, we have:
A=4,
B=-1
C=0
D=16
E=-80

Answer 4

ok

Answer 5

Now the center of the general hyperbola, are given by the subsequent formulas:
\[_C = - \frac{C}{{2A}},\quad {y_C} = - \frac{D}{{2B}}\]

Answer 6

so C=0 and y=16

Answer 7

sorry, we have:
\[{x_C} = - \frac{C}{{2A}},\quad {y_C} = - \frac{D}{{2B}}\]

Answer 8

so:
x_C=0, and y_C=8

Answer 9

oh i didnt divide by 2

Answer 10

you have to compute this:
\[{y_C} = - \frac{D}{{2B}} = - \frac{{16}}{{2 \times \left( { - 1} \right)}} = 8\]

Answer 11

8=8=-8=8???? @Michele_Laino

Answer 12

why?
we have:
\[{y_C} = - \frac{D}{{2B}} = - \frac{{16}}{{2 \times \left( { - 1} \right)}} = - \frac{{16}}{{ - 2}} = 8\]

Answer 13

oh never mind i thought you meant something else. i understand how to do the yc equation though

Answer 14

i had just forgot to divide by 2

Answer 15

we have to use the method of completion of the square, like below:

Answer 16

?

Answer 17

\[\Large \begin{gathered}
4{x^2} - {y^2} + 16y - 80 = 0 \hfill \\
4{x^2} - ({y^2} - 16y + 64 - 64) - 80 = 0 \hfill \\
4{x^2} - \left( {{y^2} - 16y + 64} \right) + 64 - 80 = 0 \hfill \\
4{x^2} - {\left( {y - 8} \right)^2} - 16 = 0 \hfill \\
\frac{{{x^2}}}{4} - \frac{{{{\left( {y - 8} \right)}^2}}}{{16}} = 1 \hfill \\
\end{gathered} \]

Answer 18

ok

Answer 19

now in order to find the x-coordinates of points A, and B:

we have to set y=8, into our equation, so we have it to solve for x