Question

Shifting Conic SectionsYou may wish to review Section 1.2 before solving Exercises 39-56.The hyperbola (y²/4) − (x²/5) = 1 is shifted 2 units down to generate the hyperbola (y + 2)²/4 − x²/5 = 1.a. Find the center, foci, vertices, and asymptotes of the new hyperbola.

Accepted Answer

Identify the original hyperbola equation: $$\frac{y^2$}{4} - \frac{x^2$}{5} = 1. $$This is a vertical hyperbola centered at the origin $$(0,0)$$ because the $$y^2$$ term is positive and comes first. Recognize the shift applied: the hyperbola is shifted 2 units down, changing $$y to$$ $$y + 2. $$The new equation is $$\frac{(y + 2)^2$}{4} - \frac{x^2$}{5} = 1. $$This means the center moves from $$(0,0) to$$ $$(0, -2). $$Find the center of the new hyperbola: since the shift is down by 2 units, the center is at $$(0, -2). $$Determine the vertices: for a vertical hyperbola of the form $$\frac{(y - k)^2$}{a^2$} - \frac{(x - h)^2$}{b^2$} = 1$$, the vertices are located $$a$$ units above and below the center. Here, $$a^2 = 4$, so a = 2$. Thus, vertices are at (0$$, -2 + 2)$$ and (0$$, -2 - 2)$$. Find the foci: the distance c$ from the center satisfies c^2$ = a^2$ + b^2$. Given a^2$ = 4$$ and $$b^2 = 5$, calculate c$$ = \sqrt{4 + 5}$$. The foci are then at (0$$, -2 + c)$$ and (0$$, -2 - c)$$. Write the equations of the asymptotes: for a vertical hyperbola centered at (h$$,k)$$, the asymptotes are given by y - k$$ = \pm \frac{a}{b}(x - h)$$. Substitute h=0$, k=-2$, a=2$, and b$$=\sqrt{5}$$ to get the asymptote equations.

Shifting Conic Sections

You may wish to review Section 1.2 before solving Exercises 39-56.

The hyperbola (y²/4) − (x²/5) = 1 is shifted 2 units down to generate the hyperbola (y + 2)²/4 − x²/5 = 1.
a. Find the center, foci, vertices, and asymptotes of the new hyperbola.

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Concetti chiave

Shifting Conic Sections

Properties of Hyperbolas

Equations of Asymptotes for Hyperbolas