How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola $(y-4)^2/16-(x+2)^2/9=1$ ?

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Answer 1 · 2018-07-02T22:55:46

Given equation of hyperbola:

$\frac{(y-4)^2}{4^2}-\frac{(x+2)^2}{3^2}=1$

Comparing above equation with standard form of vertical hyperbola: $\frac{Y^2}{a^2}-\frac{X^2}{b^2}=1$ we get

$X=x+2, Y=y-4, a=4, b=3$

$\text{eccentricity}, e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{3^2}{4^2}}=5/4$

**Center of hyperbola: **

$X=0, Y=0$

$x+2=0, y-4=0$

$x=-2, y=4$

$\text{Center}\equiv(-2, 4)$

Now, the vertices of hyper bola

$X=0, Y=\pma$

$x+2=0, y-4=\pm4$

$x=-2, y=4\pm4$

$\text {Vertices} (-2, 8)\ & \ (-2, 0)$

Now, the focii of hyper bola

$X=0, Y=\pm ae$

$x+2=0, y-4=\pm 4\cdot 5/4$

$x=-2, y=4\pm5$

$\text {focii}, (-2, 9)\ &\ (-2, -1)$

Asymptotes of hyperbola

$Y=\pma/bX$

$y-4=\pm4/3(x+2)$

$y=\pm4/3(x+2)+4$

How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola (y-4)^2/16-(x+2)^2/9=1(y-4)^2/16-(x+2)^2/9=1?