Question

Use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. y2/16−x2/36=1

Accepted Answer

Identify the standard form of the hyperbola equation. The given equation is $$\frac{y^2$}{16} - \frac{x^2$}{36} = 1$$, which matches the form $$\frac{y^2$}{a^2$} - \frac{x^2$}{b^2$} = 1. $$This means the hyperbola opens vertically (up and down). Determine the values of $$a$$ and $$b by $$comparing denominators: $$a^2 = 16$ so a = 4$, and b^2$ = 36 so$$ $$b = 6. $$The vertices are located at $$(0, \pm a)$$, which means $$(0, \pm 4). $$Find the foci using the relationship $$c^2$$ = $$a^2$$ + $$b^2. $$Calculate $$c by$$ $$c = \sqrt{16 + 36} = \sqrt{52}. $$The foci are at $$(0, \pm c)$$, or $$(0, \pm \sqrt{52}). $$Write the equations of the asymptotes. For a hyperbola of the form $$\frac{y^2$}{a^2$} - \frac{x^2$}{b^2$} = 1$$, the asymptotes are given by $$y = \pm \frac{a}{b} x. $$Substitute $$a = 4$$ and $$b = 6 to $$get $$y = \pm \frac{4}{6} x. $$Summarize the key features: vertices at $$(0, \pm 4)$$, foci at $$(0, \pm \sqrt{52})$$, and asymptotes $$y = \pm \frac{2}{3} x. $$Use these to sketch the hyperbola, plotting vertices and foci, then drawing the asymptotes as guide lines.

Use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. y²/16−x²/36=1

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Key Concepts

Standard Form of a Hyperbola

Foci of a Hyperbola

Equations of the Asymptotes