Question

Graph the hyperbola. Locate the foci and find the equations of the asymptotes. $$((x-2)^2)/25$$ - $$((y+3)^2)/16 = 1$$

Accepted Answer

Step 1: Recognize the standard form of the hyperbola. The given equation is $$ \frac{(x-2)^2}{25} - \frac{(y+3)^2}{16} = 1 $$, which matches the standard form of a hyperbola that opens horizontally: $$ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 . $$Here, $$ h = 2 $$, $$ k = -3 $$, $$ a^2 = 25 $$, and $$ b^2 = 16 . $$Step 2: Identify the center of the hyperbola. The center is given by $$ (h, k) $$, so the center is $$ (2, -3) . $$Step 3: Determine the vertices. For a horizontally opening hyperbola, the vertices are located $$ a $$ units to the left and right of the center along the x-axis. Since $$ a = \sqrt{25} = 5 $$, the vertices are $$ (2-5, -3) = (-3, -3) $$ and $$ (2+5, -3) = (7, -3) . $$Step 4: Locate the foci. The distance from the center to each focus is $$ c $$, where $$ c = \sqrt{a^2 + b^2} . $$Calculate $$ c = \sqrt{25 + 16} = \sqrt{41} . $$The foci are $$ (2-\sqrt{41}, -3) $$ and $$ (2+\sqrt{41}, -3) . $$Step 5: Find the equations of the asymptotes. For a horizontally opening hyperbola, the asymptotes are given by $$ y - k = \pm \frac{b}{a}(x - h) . $$Substituting $$ b = \sqrt{16} = 4 $$, $$ a = 5 $$, $$ h = 2 $$, and $$ k = -3 $$, the equations of the asymptotes are $$ y + 3 = \frac{4}{5}(x - 2) $$ and $$ y + 3 = -\frac{4}{5}(x - 2) .$$

Graph the hyperbola. Locate the foci and find the equations of the asymptotes. ((x-2)^2)/25 - ((y+3)^2)/16 = 1

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Key Concepts

Hyperbola Definition

Foci of a Hyperbola

Asymptotes of a Hyperbola