Question

Find the vertices and foci of the hyperbola with equation \(\dfrac{x^2}{16} - \dfrac{y^2}{48} = 1\)

Vertices: (0, +- 4); Foci: (0, +- 8)
Vertices: (0, +- 8); Foci: (0, +- 4)
Vertices: (+- 4, 0); Foci: (+- 8, 0)
Vertices: (+- 8, 0); Foci: (+- 4, 0)

Answer 1

@ganeshie8

Answer 2

straightforward question
how are you stuck on this ?

Answer 3

I know the foci is +-4, but not sure how to find the vertices.
\(c^2 = \sqrt{a^2 + b^2}\\c^2 = \sqrt{16+48}\\c^2 = \sqrt{64}\\c = 4\)

Answer 4

\[c^2 = a^2 + b^2\]

you should get \(c = \pm 8\)

Answer 5

Oh, I read it wrong,
\(c = \sqrt{a^2 + b^2}\\c = \sqrt{16+48}\\c = \sqrt{64}\\c = 8\)