Respuesta :
Answer:
Foci: (±5, 0)
Vertex: ( ±4 , 0)
Eccentricity: e = 5/4
Step-by-step explanation:
Hyperbola:
Write the equation of hyperbola in the form,
[tex]\boxed{\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1}[/tex]
9x² - 16y² = 144
Divide the entire equation by 144,
[tex]\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[/tex]
a = 4 and b =3
Eccentricity: e
[tex]\boxed{e=\sqrt{1+\dfrac{b^2}{a^2}}}[/tex]
[tex]\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}[/tex]
Foci:
(ae, 0) & (-ae , 0)
[tex]\sf \left(4*\dfrac{5}{4},0 \right) \ ; \ \left( -4*\dfrac{5}{4},0 \right)\\\\\\(5, 0) \ ; \ (-5,0)[/tex]
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)
The solution for this question is as follows:
Vertices ≈ (±4 , 0)
Foci ≈ (±5 , 0)
Eccentricity: e = 5/4
Asymptotes: y= ±(3/4)x
Hyperbola:
The location of all the points on a plane where the distance between them and two fixed points in the plane is always equal to zero is called a hyperbola.
Step by step solution:
Given: [tex]9x^{2} -16y^{2} =144[/tex]
The given equation of hyperbola can be written as:
⇒[tex]\frac{x^{2} }{16} -\frac{y^{2} }{9} =1[/tex]
On comparing with standard equation of hyperbola; [tex]\frac{x^{2} }{a} -\frac{y^{2} }{b} =1[/tex]
We get; a=4 and b=3
The eccentricity of hyperbola is given by:
9=16(e^2 - 1)
So; e^2 = (9/16) + 1
e^2 = 25/16
e = 5/4
The vertices of hyperbola are given by (±a , 0) ≈ (±4 , 0)
The foci of hyperbola are given by (±ae , 0) ≈ (±5 , 0)
The asymptotes of hyperbola are given by y= ± (b/a)x
y= ±(3/4)x
To learn more about "Hyperbola" here:
brainly.com/question/19989302
