Verified Answer :

\(\bf \cfrac{(x-{{ h}})^2}{{{ a}}^2}-\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1 \qquad center\ ({{ h}},{{ k}})\qquad vertices\ ({{ h}}\pm a, {{ k}}) \\ \quad \\\\ foci\to h\pm c\quad where\quad c=\sqrt{a^2+b^2}\to h\pm \sqrt{a^2+b^2}\\ ----------------------------\\ \textit{now, let's see your hyperbola} \\ \quad \\ \cfrac{x^2}{16}-\cfrac{y^2}{48}=1\implies \cfrac{(x-0)^2}{4^2}-\cfrac{(y-0)^2}{(\sqrt{48})^2}=1\)

and surely, you can see where the center is at, thus what h,k are,
and what "a" and "b" components are as well :)

Answer:

Vertices are (4,0),(-4,0).            

Foci are (8,0), (-8,0).    

Step-by-step explanation:

Given : The hyperbola equation \(\frac{x^2}{16}-\frac{y^2}{48}=1\)

To find : The vertices and foci of the hyperbola ?

Solution :

The general form of the hyperbola equation is

\(\cfrac{(x-{{ h}})^2}{{{ a}}^2}-\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1\)

Where, (h,k)=(0,0) is the center.

\(a^2=16\\a=4\) and \(b^2=48\\b=\sqrt{48}\)

The vertices of the hyperbola is,

\(V=(h\pm a,k)\)

Substitute the value,

\(V=(0\pm 4,0)\)

\(V=(4,0),(-4,0)\)

Therefore, Vertices are (4,0),(-4,0).

Foci is \(F=(\pm c,0)\)

Where, \(c=\sqrt{a^2+b^2}\)

\(c=\sqrt{4^2+\sqrt{48}^2}\)

\(c=\sqrt{16+48}\)

\(c=\sqrt{64}\)

\(c=8\)

\(F=(\pm 8,0)\)

Therefore, Foci are (8,0), (-8,0).

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