For the following equation of a hyperbola determine the center, vertices, foci, and asymptotes? 36x^2-y^2-24x+6y-41=0
do this just like the ellipse in your earlier post. That is, complete the squares, then review your text about hyperbolas.
36x^2-24x - (y^2-6y) = 41
36(x - 1/3)^2 - (y-3)^2 = 41 + 36/9 - 9
(x-1/3)^2 - (y-3)^2/36 = 1
To find the center, vertices, foci, and asymptotes of a hyperbola, we first need to put the equation in the standard form of a hyperbola:
\[ \frac{{(x-h)^2}}{{a^2}} - \frac{{(y-k)^2}}{{b^2}} = 1 \]
or
\[ \frac{{(y-k)^2}}{{b^2}} - \frac{{(x-h)^2}}{{a^2}} = 1 \]
where (h, k) represents the coordinates of the center, and a and b are the distances from the center to the vertices.
Let's start by rearranging the given equation into the standard form:
\[ 36x^2 - y^2 - 24x + 6y - 41 = 0 \]
Taking all the x terms together and the y terms together:
\[ 36x^2 - 24x - y^2 + 6y = 41 \]
Now, we complete the square for both the x and y terms separately:
For the x terms:
\[ 36(x^2 - \frac{2}{3}x) \]
To complete the square, we need to add the square of half the coefficient of the x term:
\[ 36(x^2 - \frac{2}{3}x + (\frac{1}{3})^2) \]
Simplifying:
\[ 36(x^2 - \frac{2}{3}x + \frac{1}{9}) \]
\[ 36(x - \frac{1}{3})^2 \]
For the y terms:
\[ -(y^2 - 6y) \]
To complete the square, we add the square of half the coefficient of the y term:
\[ -(y^2 - 6y + 9) \]
Simplifying:
\[ -(y - 3)^2 \]
Rearranging the equation:
\[ 36(x - \frac{1}{3})^2 - (y - 3)^2 = 41 \]
Dividing both sides by 41 to get 1 on the right-hand side:
\[ \frac{(x - \frac{1}{3})^2}{\frac{41}{36}} - \frac{(y - 3)^2}{41} = 1 \]
Comparing this form to the standard form, we can see that:
\[ h = \frac{1}{3} \]
\[ k = 3 \]
\[ a^2 = \frac{41}{36} \]
\[ b^2 = 41 \]
The center of the hyperbola is at (h, k) = (1/3, 3).
To find the vertices, we need to find a. Taking the square root of a^2:
\[ a = \sqrt{\frac{41}{36}} = \frac{\sqrt{41}}{6} \]
The distance from the center to the vertices is a. So the vertices are located at:
\[ (1/3 + a, 3) \] and \[ (1/3 - a, 3) \]
Now, let's find the foci. The distance from the center to the foci is c, where c is calculated using the equation:
\[ c^2 = a^2 + b^2 \]
\[ c^2 = \frac{41}{36} + 41 \]
\[ c^2 = \frac{1}{36}(41 + (36)(41)) \]
\[ c^2 = \frac{37}{36}(41) \]
Taking the square root:
\[ c = \frac{\sqrt{37 \times 41}}{6} \]
The foci are located at:
\[ (1/3 + c, 3) \] and \[ (1/3 - c, 3) \]
Finally, let's find the asymptotes. The equations of the asymptotes are of the form:
\[ y = mx + b \] and \[ y = -mx + b \]
where m is the slope of the asymptote and b is a constant.
The slopes of the asymptotes are given by:
\[ m = \pm \frac{b}{a} \]
In this case, the formula gives us:
\[ m = \pm \frac{41}{\sqrt{41}/6} = \pm \frac{6 \times 41}{\sqrt{41}} = \pm 6\sqrt{41} \]
To find b, let's substitute the coordinates of the center (h, k) into the equation:
\[ b = y - mx \]
\[ 3 = (6\sqrt{41})(1/3) \]
\[ b = 3 - 2\sqrt{41} \]
Therefore, the two asymptotes are:
\[ y = \pm (6\sqrt{41})x + (3 - 2\sqrt{41}) \]
In summary, for the given equation \( 36x^2 - y^2 - 24x + 6y - 41 = 0 \):
- The center is (1/3, 3).
- The vertices are (1/3 + a, 3) and (1/3 - a, 3), where a = sqrt(41)/6.
- The foci are (1/3 + c, 3) and (1/3 - c, 3), where c = sqrt(37 * 41)/6.
- The asymptotes are y = (6sqrt(41))x + (3 - 2sqrt(41)) and y = -(6sqrt(41))x + (3 - 2sqrt(41)).