The center of a hyperbola is located at (0, 0). One focus is located at (0, 5) and its associated directrix is represented by th
2 answers:
For a hyperbola 
where
the directrix is the line
and the focus is at (0, c).
Here, we have c = 5, a² = 9, so b² = 5² - 9 = 16.
a = √9 = 3
b = √16 = 4
Your hyperbola's constants are ...
a = 3
b = 4
______
Please note that the equation of a hyperbola has a negative sign for one of the terms. The equation given in your problem statement is that of an ellipse.
where
the directrix is the line
and the focus is at (0, c).
Here, we have c = 5, a² = 9, so b² = 5² - 9 = 16.
a = √9 = 3
b = √16 = 4
Your hyperbola's constants are ...
a = 3
b = 4
______
Please note that the equation of a hyperbola has a negative sign for one of the terms. The equation given in your problem statement is that of an ellipse.
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Answer:
Step-by-step explanation:
We'll just work on solving both so you can see what's involved in solving an absolute value equation. Because an absolute value is a distance, we can have that distance being both to the right on the number line of the number in question or to the left. For example, from 2 on the number line, the numbers that are 5 units away are 7 and -3. Using that logic, we will simplify the equation down so we can set up the 2 basic equations needed to solve for x.
If then
What you need to remember here is that you cannot distribute into a set of absolute values like you would a set of parenthesis. The -2 needs to be divided away:
Now we can set up the 2 main equations for this which are
.5x + 1.5 = .5 and .5x + 1.5 = -.5
Knowing that an absolute value will never equal a negative number (because absolute values are distances and distances will NEVER be negative), once we remove the absolute value signs we can in fact state that the expression on the left can be equal to a negative number on the right, like in the second equation above.
Solving the first one:
.5x = -1 so
x = -2
Solving the second one:
.5x = -2 so
x = -4