Answer:
Answer:
Foci: (±5, 0)
Vertex: ( ±4 , 0)
Eccentricity: e = 5/4
Explanation:
Hyperbola:
Write the equation of hyperbola in the form,
\boxed{\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1}
a
2
x
2
−
b
2
y
2
=1
9x² - 16y² = 144
Divide the entire equation by 144,
\begin{gathered}\sf \dfrac{9x^2}{144}-\dfrac{16y^2}{144}=\dfrac{144}{144}\\\\\\ \dfrac{x^2}{16}-\dfrac{y^2}{9}=1\\\\\\\dfrac{x^2}{4^2}-\dfrac{y^2}{3^2}=1\end{gathered}
144
9x
2
−
144
16y
2
=
144
144
16
x
2
−
9
y
2
=1
4
2
x
2
−
3
2
y
2
=1
a = 4 and b =3
Eccentricity: e
\boxed{e=\sqrt{1+\dfrac{b^2}{a^2}}}
e=
1+
a
2
b
2
\begin{gathered}\sf = \sqrt{1+\dfrac{9}{16}}\\\\=\sqrt{\dfrac{16}{16}+\dfrac{9}{16}}\\\\=\sqrt{\dfrac{25}{16}}=\sqrt{\dfrac{5*5}{4*4}}\\\\\\=\dfrac{5}{4}\end{gathered}
=
1+
16
9
=
16
16
+
16
9
=
16
25
=
4∗4
5∗5
=
4
5
Foci:
(ae, 0) & (-ae , 0)
\begin{gathered}\sf \left(4*\dfrac{5}{4},0 \right) \ ; \ \left( -4*\dfrac{5}{4},0 \right)\\\\\\(5, 0) \ ; \ (-5,0)\end{gathered}
(4∗
4
5
,0) ; (−4∗
4
5
,0)
(5,0) ; (−5,0)
Vertex of hyperbola:
The vertex of the hyperbola is a point where the hyperbola cuts the axis. (-a , 0) ; (a , 0)
Vertex : (-4 , 0) ; (4 , 0)