Question

find the standard form of the equation of the hyperbola shown.
the standard form of the equation is \\(\square\\).

Explanation:

Step1: Identify the center

The center of the hyperbola is at \((h, k)\). From the graph, the center is at \((0, 0)\) (since the midpoint of the vertices and the intersection of the asymptotes is at the origin).

Step2: Determine the orientation and \(a\), \(b\)

This is a vertical hyperbola (opens up and down) because the vertices are on the \(y\)-axis. The distance from the center to a vertex (\(a\)): the vertices are at \((0, 5)\) and \((0, -5)\)? Wait, no, looking at the graph, the vertices seem to be at \((0, 5)\) and \((0, -5)\)? Wait, no, the dashed rectangle: from \(x = -6\) to \(x = 6\) (so length \(12\), so \(b = 6\)) and \(y = -5\) to \(y = 5\)? Wait, no, the vertices are at \((0, 5)\) and \((0, -5)\)? Wait, the standard form for a vertical hyperbola is \(\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\).

Wait, the center is \((0,0)\). The vertices: looking at the graph, the top vertex is at \((0, 5)\) and bottom at \((0, -5)\), so \(a = 5\)? Wait, no, the dashed rectangle: from \(x = -6\) to \(x = 6\) (so the horizontal distance from center to the side of the rectangle is \(b = 6\)) and \(y = -5\) to \(y = 5\) (vertical distance \(a = 5\))? Wait, no, the asymptotes for a vertical hyperbola are \(y = \pm \frac{a}{b}x\). Wait, the asymptotes here: let's find the slope. The asymptotes pass through \((-6, 5)\) and \((6, -5)\)? Wait, no, the dashed lines: the asymptotes go through \((-6, 5)\) and \((6, -5)\)? Wait, the slope of the asymptote: from \((0,0)\) to \((6, 5)\)? Wait, no, the dashed rectangle has corners at \((-6, 5)\), \((6, 5)\), \((6, -5)\), \((-6, -5)\). So the length of the horizontal side of the rectangle is \(12\) (from \(x=-6\) to \(x=6\)), so \(b = 6\) (since \(b\) is the distance from center to the side of the rectangle horizontally). The vertical side is \(10\) (from \(y=-5\) to \(y=5\)), so \(a = 5\) (distance from center to vertex vertically).

Step3: Write the standard form

For a vertical hyperbola with center \((h, k) = (0,0)\), the standard form is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Here, \(a = 5\) (so \(a^2 = 25\)) and \(b = 6\) (so \(b^2 = 36\)). Wait, no, wait: the vertices are at \((0, 5)\) and \((0, -5)\), so \(a = 5\). The asymptotes: the slope is \(\pm \frac{a}{b}\). From the rectangle, the horizontal length is \(2b = 12\) (from \(x=-6\) to \(x=6\)), so \(b = 6\). So the standard form is \(\frac{y^2}{25} - \frac{x^2}{36} = 1\)? Wait, no, wait: if the hyperbola is vertical, the standard form is \(\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\). Center \((0,0)\), so \(h = 0\), \(k = 0\). \(a\) is the distance from center to vertex: the vertices are at \((0, 5)\) and \((0, -5)\), so \(a = 5\), \(a^2 = 25\). \(b\) is the distance from center to the side of the rectangle horizontally: the rectangle goes from \(x = -6\) to \(x = 6\), so \(b = 6\), \(b^2 = 36\). So the equation is \(\frac{y^2}{25} - \frac{x^2}{36} = 1\)? Wait, but let's check the asymptotes. The asymptotes for a vertical hyperbola are \(y = \pm \frac{a}{b}x = \pm \frac{5}{6}x\). Let's see if the asymptotes pass through \((6, 5)\): \(y = \frac{5}{6}(6) = 5\), yes! So that's correct. So the standard form is \(\frac{y^2}{25} - \frac{x^2}{36} = 1\).

Answer:

\(\frac{y^2}{25} - \frac{x^2}{36} = 1\)