Respuesta :

[tex]\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[/tex]

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 [tex]\frac{x^2}{16}-\frac{y^2}{48}=1[/tex]

To find : The vertices and foci of the hyperbola ?

Solution :

The general form of the hyperbola equation is

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

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

[tex]a^2=16\\a=4[/tex] and [tex]b^2=48\\b=\sqrt{48}[/tex]

The vertices of the hyperbola is,

[tex]V=(h\pm a,k)[/tex]

Substitute the value,

[tex]V=(0\pm 4,0)[/tex]

[tex]V=(4,0),(-4,0)[/tex]

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

Foci is [tex]F=(\pm c,0)[/tex]

Where, [tex]c=\sqrt{a^2+b^2}[/tex]

[tex]c=\sqrt{4^2+\sqrt{48}^2}[/tex]

[tex]c=\sqrt{16+48}[/tex]

[tex]c=\sqrt{64}[/tex]

[tex]c=8[/tex]

[tex]F=(\pm 8,0)[/tex]

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

Otras preguntas