Graph the equation. Identify the vertices, foci, and asymptotes of the hyperbola.
Y^2/16-x^2/36=1
I think....vertices (0,4),(0,-4) foci???
asymptotes y=+/- 2x/3
To graph the equation of a hyperbola, you can follow these steps:
1. Rewrite the equation in standard form: \(\frac{{y^2}}{{a^2}} - \frac{{x^2}}{{b^2}} = 1\), where \(a\) and \(b\) are positive real numbers.
Comparing the given equation \(\frac{{y^2}}{{16}} - \frac{{x^2}}{{36}} = 1\) with the standard form, we have \(a^2 = 16\) and \(b^2 = 36\). Taking the square root of both sides, we get \(a = 4\) and \(b = 6\).
2. Identify the center of the hyperbola.
The center of the hyperbola is given by the coordinates \((h, k)\), where \(h\) is the x-coordinate and \(k\) is the y-coordinate. In this equation, the center is at the origin (0, 0).
3. Plot the vertices.
The vertices are points on the transverse axis, which is the line passing through the center. The distance from the center to each vertex is \(a\). Since \(a = 4\), the vertices will be located at (0, 4) and (0, -4).
4. Plot the foci.
The foci are points along the transverse axis, which is the line passing through the center. The distance from the center to each focus is denoted by \(c\). To find the value of \(c\), use the equation \(c^2 = a^2 + b^2\). Plugging in the values, we have \(c^2 = 16 + 36 = 52\), and taking the square root on both sides gives \(c = \sqrt{52} = 2\sqrt{13}\). Therefore, the foci will be located at (0, \(2\sqrt{13}\)) and (0, \(-2\sqrt{13}\)).
5. Plot the asymptotes.
The equations of the asymptotes in the standard form for a hyperbola are \(y = \pm \frac{b}{a}x + k\), where \(k\) is the y-coordinate of the center. In this case, the asymptotes will be \(y = \pm \frac{2}{3}x\) since \(k = 0\).
Now, you can plot the vertices (0, 4) and (0, -4), the foci (0, \(2\sqrt{13}\)) and (0, \(-2\sqrt{13}\)), and the asymptotes \(y = \frac{2}{3}x\) and \(y = -\frac{2}{3}x\).