vertice (0,e) and (6,2) asymptotes y = ax and y=4-22 The Standard form of Hyperbola is; with horrential tranverge asy. (x-4) ²- (y-x)² = 6² in the given vertices verticy of the the I Coordinate of both the hyperbola are same, it meary that Hyperbola hay horizontal tramvyverse axis. verticey are (hta, k) Cluation of asymptotes, F y=k+b(x-h) and y=k-b(x-h) "the mild point of the a : Center is "vertice); Center (hak) = (0+62+2) (²) Solution (h, k) = (3,2) The distance between the verticy is qual to 29, 2a= 6-0 a=3 Comparing the equations of asymptoty and Ⓒ, b=2 valuy of (h.K) (2-3)² 3² 3 and → b put in quation 1, (y-2)² = 1
vertice (0,e) and (6,2) asymptotes y = ax and y=4-22 The Standard form of Hyperbola is; with horrential tranverge asy. (x-4) ²- (y-x)² = 6² in the given vertices verticy of the the I Coordinate of both the hyperbola are same, it meary that Hyperbola hay horizontal tramvyverse axis. verticey are (hta, k) Cluation of asymptotes, F y=k+b(x-h) and y=k-b(x-h) "the mild point of the a : Center is "vertice); Center (hak) = (0+62+2) (²) Solution (h, k) = (3,2) The distance between the verticy is qual to 29, 2a= 6-0 a=3 Comparing the equations of asymptoty and Ⓒ, b=2 valuy of (h.K) (2-3)² 3² 3 and → b put in quation 1, (y-2)² = 1
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section11.3: Hyperbolas
Problem 36E
Related questions
Question
Explain it to me. How do you came up with 2
![vertice (0,2) and (6,2)
asymptotes y = 2²
The Standard form of Hyperbola is; with horizential.
tranverge aus.
in the
vertices of the
the
(x-7) ²_ (y-x)² =1
-
9²
6²
given verticy
I Coordinaty of both the
hyperbola are Jame, it meary.
that
and y= 4-2
Solution
Hyperbola hay horizontal tranyverse axis.
vertice are (hta, k)
Cluation of asymptotes,
.
g=k+ b (x-h) and y=k-b(x-h)
"the midpoint of the vertices;
: Center is
Center (tok) = (0+ 6, 2+2).
2a= 6-0
2a=6
a=3
(h.k)= (3,2)
The distance between the verticy is qual to 29,
(2-3) ²
32
3
Comparing the equations of asymplatey and Ⓡ,
+ 6 = +3
2
b=2
valuy of (h.K)
3
a
and
2
b put in quation 1,
(y-2)² = 1
2
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbef53a0e-143c-440f-8b26-4b6191061c47%2F79d803e2-acf6-4169-b4a2-4bb233f4e02e%2Fqn5vv1_processed.jpeg&w=3840&q=75)
Transcribed Image Text:vertice (0,2) and (6,2)
asymptotes y = 2²
The Standard form of Hyperbola is; with horizential.
tranverge aus.
in the
vertices of the
the
(x-7) ²_ (y-x)² =1
-
9²
6²
given verticy
I Coordinaty of both the
hyperbola are Jame, it meary.
that
and y= 4-2
Solution
Hyperbola hay horizontal tranyverse axis.
vertice are (hta, k)
Cluation of asymptotes,
.
g=k+ b (x-h) and y=k-b(x-h)
"the midpoint of the vertices;
: Center is
Center (tok) = (0+ 6, 2+2).
2a= 6-0
2a=6
a=3
(h.k)= (3,2)
The distance between the verticy is qual to 29,
(2-3) ²
32
3
Comparing the equations of asymplatey and Ⓡ,
+ 6 = +3
2
b=2
valuy of (h.K)
3
a
and
2
b put in quation 1,
(y-2)² = 1
2
2
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