Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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TRANSCRIBE THE FOLLOWING TEXT IN DIGITAL FORMAT
Transcribed Image Text:Given Conic Equation
9x² - 4y² +54x +16y +29 = 0
9x² - 4y² + 54x + 1by = -29
Complete the square for 9x²+54x
foome 92²+bx+c to find.
Use the
a, b, and c
a=9
6=54
Consider the vertex form of parabola
a(x+d)² + e
Find the value
da b
29
2
54
2x9
b2
पव
Value
CHO
of d using the
=
Find the value of e using the formula
е
e= c-
20-(54)2
value of e
9(2+3)²-81
= -81
4x9
Subtitute the values of a, dande in
the vertex
of d = 5/20₁
b
29
1/2
2
Subtitute 9(2+3)²-81 for 9x² +54x
in the equation
to find the value of
9x² - 4y² +54x + 16y=-29
9(2+3) ²2²-81-4y² + 16y = -
9 (x+3)² - 4y²+1by
9(x+3)² - 4y² +1by
C=O
= C-
= 3
Complete the Square for _4y² +167
Use the form ax²+bx+c, to find the value of a, band c
a = - 4
b=16
Consider the vertex form of parabola
G (a+d)² + e
=
e formula
52
49
16
2X (-4)
= -2
=-29+8)
= +62
(16)2
= 16
4 X (-4)
Subtitute the value of a, d ande into the vertex
form
-4(y-2)² +16
Subtitute -4 (9-2)² +16 for - 4y² +16y in the equation.
Gx² - 4y² + 54x+16y=-29
9 (x+3)² - 4(y + 2)² + 16 = -29+8|
Transcribed Image Text:9 (x+3)² - 4(4+2)²
9(2+3)² -4(y+2)² = 36
divide each term by 36 to make the
right side equal to one.
9 (x+3)²
36
(2+3)²
4
1
(x-4)²
92
4(५+2)2
36
(y-2)2
9
This is the form of a hyperbola.
Standrad form of hyperbolo
(Y-K 2²
b2
= -29 +81-16
Here, a=2, b= 3, K=2, h=-3
The center of hyperbola (hik)
(-3,2)
Find the distance from the center to a focus
<= √6²45²
= √(²)² + (3)² = √√4+9 =√13
Find the eccentricity
eccentricity =
Find the focal parameter
Focal Parameter
36
= 1
Find the vertices
The first vertex of a hyperbola can be found by
adding a to h
(h+a, k)
(-1,2)
19²+52
= 1
The second vestex of a hyperbola can be by found
by Subtracting a from h
(n-a, k)
36
(-5,2)
The vertices of a hyperbola (-1,2), (-5,2)
Find the foci
The first focus of a hyperbola can be found by
adding (toh
(h+(₂ k)
(-3+√√13,2)
Seond second focus of a hyperbola
found by Subtracting c from h
(h-C, K) (-3-√√13, 2)
The foci of a hyperbola (-3+√73, 2) (-3-√73,2)
First Asymptotes
y = 3x + 13
2
2
Second Asymptates
y = ± 3 (x-(-3)+2
2
√√/2²+32
2
62
=32
√a²+5²
113
Asymptotes follow the form y: I bla-h)
+
because this hyperbola open left and right
a
y=-321-57/2
can be
2
=9/13
13
+K
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