What are the foci of the hyperbola with equation 16y^2-9x^2-144?
AAAaannndd the bot gets it wrong yet again!
all that typesetting gibberish just makes things unreadable to us poor humans. Not only that, but the transverse axis is vertical.
16y^2-9x^2 = 144
y^2/9 - x^2/16 = 1
So a=3 and b=4, making c=5
The foci are at (0,±5)
I apologize for the error in my previous response. You are correct that the transverse axis is vertical and the foci are at (0,±5). Thank you for bringing this to my attention.
To identify the foci of a hyperbola given its equation, we need to ensure that the equation is in a standard form. The standard form of a hyperbola centered at the origin (0, 0) is:
(x^2 / a^2) - (y^2 / b^2) = 1 (if the horizontal axis is the major axis)
or
(y^2 / a^2) - (x^2 / b^2) = 1 (if the vertical axis is the major axis)
In your equation, 16y^2 - 9x^2 = 144, let's first divide both sides by 144 to make the right side equal to 1:
(16y^2 / 144) - (9x^2 / 144) = 144 / 144
y^2 / 9 - x^2 / 16 = 1
Now, we can see that the value of a^2 is 16, and b^2 is 9.
The distance c, which represents the distance between the center and each focus, can be calculated using the equation: c^2 = a^2 + b^2.
For our hyperbola equation, we have: c^2 = 16 + 9 = 25.
To find c, we take the square root of both sides: c = √25 = 5.
Now we can determine the foci by finding the values (h ± c, k), where (h, k) represents the center of the hyperbola.
In this case, since the center is (0, 0), the foci will be (0 ± 5, 0) or simply (±5, 0).
Therefore, the foci of the hyperbola with equation 16y^2 - 9x^2 = 144 are (5, 0) and (-5, 0).
We can start by putting the equation into standard form:
\[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\]
where $(h,k)$ is the center of the hyperbola. To get there, we first need to move the constant to the other side:
\[16y^2 - 9x^2 = 144\]
Dividing both sides by 144, we have:
\[\frac{y^2}{9} - \frac{x^2}{16} = 1\]
So the center is at $(h,k) = (0,0)$, and $a=3$, $b=4$. Therefore, the distance from the center to each focus is:
\[c = \sqrt{a^2 + b^2} = 5\]
The foci are located along the transverse axis, which is the x-axis in this case. So the coordinates of the foci are $(\pm c, 0)$, giving us a final answer of:
\[(\pm 5, 0)\]