Find the center, foci, vertices and asymptotes of the hyperbola $\displaystyle y^2 = 16x - 8$. Sketch its graph.
We can rewrite the equation as $\displaystyle y^2 = 16 \left(x - \frac{1}{2} \right)$. This parabola opens to the right with vertex at $\displaystyle \left( \frac{1}{2}, 0 \right)$. It is obtain from the parabola $y^2 = 16x$ by shifting its $\displaystyle \frac{1}{2}$ units to the right. Since $4p = 16$, we have $\displaystyle p = 4$. So the focus is $\displaystyle 4$ units from the left of the vertex and the directrix is $\displaystyle 4$ units from the right of the vertex.
Therefore, the focus is at
$\displaystyle \left( \frac{1}{2}, 0 \right) \to \left( \frac{1}{2} + 4, 0 \right) = \left( \frac{9}{2}, 0 \right)$
and the directrix is the line
$\displaystyle x = \frac{1}{2} - 4 = \frac{-7}{2}$
Monday, November 16, 2015
College Algebra, Chapter 8, 8.4, Section 8.4, Problem 12
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