Question

Graph the hyperbola. Locate the foci and find the equations of the asymptotes. $$4y^2$$ - $$x^2 = 16$$

Accepted Answer

Rewrite the given equation in standard form for a hyperbola. Divide through by 16 to normalize the equation: $$ \frac{4y^2}{16} - \frac{x^2}{16} = 1 $$, which simplifies to $$ \frac{y^2}{4} - \frac{x^2}{16} = 1 . $$Identify the type of hyperbola. Since the $$ y^2 $$ term is positive and comes first, this is a vertical hyperbola. The standard form for a vertical hyperbola is $$ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 . $$Here, $$ a^2 = 4 $$ and $$ b^2 = 16 $$, so $$ a = 2 $$ and $$ b = 4 . $$Determine the foci. The distance from the center to each focus is given by $$ c = \sqrt{a^2 + b^2} . $$Substitute $$ a^2 = 4 $$ and $$ b^2 = 16 $$ into the formula to find $$ c . $$The foci are located at $$ (0, \pm c) $$ since the hyperbola is vertical. Find the equations of the asymptotes. For a vertical hyperbola, the asymptotes are given by $$ y = \pm \frac{a}{b}x . $$Substitute $$ a = 2 $$ and $$ b = 4 $$ into the formula to write the equations of the asymptotes. Sketch the graph. Plot the center at $$ (0, 0) $$, draw the vertices at $$ (0, \pm a) $$, and sketch the asymptotes as diagonal lines passing through the center. Use the foci and the asymptotes to guide the shape of the hyperbola.

Graph the hyperbola. Locate the foci and find the equations of the asymptotes. 4y^2 - x^2 = 16

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Key Concepts

Hyperbola Definition

Foci of a Hyperbola

Asymptotes of a Hyperbola