Question

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`?

Answer 1

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`