AIOU Solved Assignment Code 1309 Mathematics III 2023 Assignment 2.pptx

ALLAMA IQBAL OPEN UNIVERSITY ISLAMABAD
(Department of Mathematics)
Course: Mathematics-III (1309)
Semester: Spring, 2023
Level: Intermediate
ASSIGNMENT No. 2
Q.1 (a)Find equation of hyperbola with vertex (–1, 7) and asymptotes
y–5 = ± 8(x + 1)
(b) Analyze the conic 3(x+2)2 + 4 (y + 1)2 = 12
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(a) To find the equation of a hyperbola with vertex (h, k) and
asymptotes given by the equations y - k = ±(a/b)(x - h), where a is the
distance from the center to the vertices, and b is the distance from the
center to the foci, we can use the standard form equation of a
hyperbola:
(x - h)²/a² - (y - k)²/b² = 1
Given that the vertex is (-1, 7) and the asymptotes are y - 5 = ±8(x + 1), we
can determine the values of h, k, a, and b.
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Comparing the equation of the asymptotes with the standard form equation,
we have:
h = -1
k = 7
a/b = 8
From a/b = 8, we can rewrite it as a = 8b.
The distance from the center to the vertices is a, so a = 8b implies that 8b =
a.
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To determine the value of b, we need to find the distance from the center to
the asymptote, which is the perpendicular distance from the center to either
of the asymptotes. In this case, the distance is 8.
Using the distance formula for a point and a line, the distance from the point
(-1, 7) to the line y - 5 = 8(x + 1) is:
d = |(-8(-1) + 1(7) - 5)|/√(8² + 1²) = 8/√65
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Since this distance is the same as the distance from the center to the
asymptote (b), we can equate them:
8/√65 = b
Now that we have the values of h, k, a, and b, we can substitute them into
the standard form equation to obtain the equation of the hyperbola:
(x + 1)²/(8/√65)² - (y - 7)²/(8/√65)² = 1
Simplifying the equation, we get:
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(x + 1)²/64/65 - (y - 7)²/64/65 = 1
Multiplying through by 64/65, we obtain the final equation of the hyperbola:
65(x + 1)² - 65(y - 7)² = 64
(b) The given equation is 3(x + 2)² + 4(y + 1)² = 12.
To analyze this equation, we first need to bring it into the standard form of a
conic section.
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Dividing the entire equation by 12, we have:
(x + 2)²/4 + (y + 1)²/3 = 1
Comparing this equation to the standard form of an ellipse:
(x - h)²/a² + (y - k)²/b² = 1
We can determine the values of h, k, a, and b:
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h = -2
k = -1
a² = 4
b² = 3
The center of the conic section is given by (h, k) = (-2, -1).
The major axis of the ellipse is the x-axis because a² = 4 is greater than b² =
3.
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The length of the major axis is 2a, so the length of the major axis is 2 X √(4)
= 4.
The minor axis of the ellipse is the y-axis because b² = 3 is smaller than a² =
4.
The length of the minor axis is 2b, so the length of the minor axis is 2 X √(3)
≈ 3.464.
Therefore, the conic section is an ellipse centered at (-2, -1), with the major
axis along the x-axis and a length of 4, and the minor axis along the y-axis
with a length of approximately 3.464.
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Q. 2 (a)Find equation of hyper bola having vertices (0, 6) and (6, 6)
foci 10 units apart.
(b) Find equation of tangent to the circle x2 + y2 + 28x + 5y + 71 = 0
at (2, 296)
(a) To find the equation of a hyperbola given the vertices and foci, we can
use the standard form equation of a hyperbola:
(x - h)²/a² - (y - k)²/b² = 1
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Given that the vertices are (0, 6) and (6, 6), we can determine the values of
h, k, a, and b.
The midpoint between the vertices gives us the center of the hyperbola,
which is (h, k). So the center is ((0 + 6)/2, (6 + 6)/2) = (3, 6).
The distance between the vertices is 6 units, which is equal to 2a, so a = 3.
The distance between the foci is 10 units. Since the foci lie on the transverse
axis (the line passing through the vertices), the distance between the foci is
2ae, where e is the eccentricity of the hyperbola.
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2ae = 10
Substituting a = 3:
2 X 3 X e = 10
6e = 10
e = 10/6 = 5/3
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Now that we have the values of h, k, a, and e, we can substitute them into the
standard form equation to obtain the equation of the hyperbola:
(x - 3)²/3² - (y - 6)²/b² = 1
Simplifying, we have:
(x - 3)²/9 - (y - 6)²/b² = 1
Since e = 5/3, we can also determine b using the relationship:
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b² = a²(1 + e²)
b² = 3²(1 + (5/3)²) = 9(1 + 25/9) = 9(34/9) = 34
Substituting b² = 34 into the equation, we get:
(x - 3)²/9 - (y - 6)²/34 = 1
Therefore, the equation of the hyperbola is:
(x - 3)²/9 - (y - 6)²/34 = 1
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(b) To find the equation of a tangent to a circle, we need to find the slope of
the tangent at the given point (2, 296) on the circle.
The given equation of the circle is:
x² + y² + 28x + 5y + 71 = 0
To find the slope of the tangent, we need to find the derivative of the
equation with respect to x.
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Differentiating both sides of the equation:
2x + 2yy' + 28 + 5y' = 0
Simplifying:
2x + 2yy' + 5y' = -28
Grouping the y' terms:
(2y + 5)y' = -2x - 28
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Dividing by (2y + 5):
y' = (-2x - 28)/(2y + 5)
Now, substituting the given point (2, 296) into the equation:
y' = (-2(2) - 28)/(2(296) + 5) = (-4 - 28)/(592 + 5) = -32/597
Therefore, the slope of the tangent at (2, 296) is -32/597.
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Using the point-slope form of a line (y - y₁) = m(x - x₁), where (x₁, y₁) is the
given point and m is the slope, we can write the equation of the tangent:
(y - 296) = (-32/597)(x - 2)
y - 296 = (-32/597)x + (64/597)
Bringing all terms to one side, we get:
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(-32/597)x + y - (64/597) - 296 = 0
Multiplying through by 597 to eliminate fractions, we obtain the final
equation of the tangent:
-32x + 597y - 360 - 188032 = 0
-32x + 597y - 188392 = 0
Therefore, the equation of the tangent to the circle x² + y² + 28x + 5y + 71 =
0 at the point (2, 296) is:
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-32x + 597y - 188392 = 0
Q. 3 (a)Find equation for the ellipse with foci (0, ± 2) and major axis
with ends (0, ± 4)
(b)Sketch the parabola y = 4x2 + 8x + 5, point out focus and vertex.
(a) To find the equation of an ellipse given the foci and major axis endpoints,
we can use the standard form equation of an ellipse:
(x - h)²/a² + (y - k)²/b² = 1
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Given that the foci are (0, ±2) and the major axis endpoints are (0, ±4), we
can determine the values of h, k, a, and b.
The center of the ellipse is the midpoint between the major axis endpoints,
which is (h, k). So the center is ((0 + 0)/2, (4 + (-4))/2) = (0, 0).
The distance from the center to each focus is c, and the distance from the
center to each major axis endpoint is a.
From the given information, we have:
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c = 2
a = 4
We can determine b using the relationship:
c² = a² - b²
(2)² = (4)² - b²
4 = 16 - b²
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b² = 16 - 4
b² = 12
Now that we have the values of h, k, a, and b, we can substitute them into
the standard form equation to obtain the equation of the ellipse:
(x - 0)²/4² + (y - 0)²/√12² = 1
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x²/16 + y²/12 = 1
Therefore, the equation of the ellipse is:
x²/16 + y²/12 = 1
(b) To sketch the parabola y = 4x² + 8x + 5 and identify the focus and vertex,
we can analyze the equation.
The given equation is in the form y = ax² + bx + c, where a = 4, b = 8, and c
= 5.
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Comparing the equation to the standard form of a parabola y = a(x - h)² + k,
we can determine the values of h and k.
To find the vertex, we can use the formula:
h = -b/2a
h = -8/(2 X 4) = -8/8 = -1
Substituting h = -1 into the equation, we can find k:
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k = 4(-1)² + 8(-1) + 5 = 4 - 8 + 5 = 1
Therefore, the vertex of the parabola is (-1, 1).
To find the focus, we can use the formula:
F = (h, k + 1/(4a))
F = (-1, 1 + 1/(4 X 4))
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F = (-1, 1 + 1/16)
F = (-1, 17/16)
Therefore, the focus of the parabola is (-1, 17/16).
Sketching the parabola using the vertex and focus information, we can plot
the points (-1, 1) and (-1, 17/16) and draw a curve that opens upwards.
Q. 4 (a)Find a vector that is oppositely directed to v = [3, –4] and half
the length of v
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(b) Find initial point of u = [–3, 1, 2], if the terminal point is (5, 0, –1).
(a) To find a vector that is oppositely directed to v = [3, -4] and half the
length of v, we can multiply v by -1 and divide it by 2.
The vector that is oppositely directed to v is -v = -[3, -4] = [-3, 4].
To obtain a vector that is half the length of v, we can multiply v by 1/2:
(1/2)v = (1/2)[3, -4] = [3/2, -2].
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Therefore, a vector that is oppositely directed to v = [3, -4] and half the
length of v is [-3/2, 2].
(b) To find the initial point of u = [-3, 1, 2], given the terminal point is (5,
0, -1), we need to find the difference between the terminal point and the
vector u.
The initial point of u can be calculated by subtracting u from the terminal
point:
Initial point = Terminal point - u
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Initial point = (5, 0, -1) - (-3, 1, 2)
Initial point = (5 + 3, 0 - 1, -1 - 2)
Initial point = (8, -1, -3)
Therefore, the initial point of u = [-3, 1, 2], with the terminal point (5, 0, -1),
is (8, -1, -3).
Q. 5 (a)Find direction cosines and directional angles of the vector: 2i–
4j+4k
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(b) Find ‘a’ vector perpendicular to vectors: u = [2, –1, 3], v = [–1,3,1]
(a) To find the direction cosines of a vector, we divide each component of
the vector by its magnitude. The magnitude of a vector v = [a, b, c] can be
calculated as:
|v| = √(a² + b² + c²)
For the given vector v = 2i - 4j + 4k, the magnitude is:
|v| = √((2)² + (-4)² + (4)²) = √(4 + 16 + 16) = √36 = 6
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Now, we can calculate the direction cosines. The direction cosines of a
vector v = [a, b, c] are given by the ratios:
cos α = a/|v|
cos β = b/|v|
cos γ = c/|v|
Therefore, for v = 2i - 4j + 4k:
cos α = 2/6 = 1/3
cos β = -4/6 = -2/3
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cos γ = 4/6 = 2/3
To find the directional angles, we can take the inverse cosine (arccos) of the
direction cosines:
α = arccos(1/3)
β = arccos(-2/3)
γ = arccos(2/3)
Note that the directional angles α, β, and γ are measured in radians.
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(b) To find a vector perpendicular to vectors u = [2, -1, 3] and v = [-1, 3, 1],
we can take their cross product.
The cross product of two vectors u = [u₁, u₂, u₃] and v = [v₁, v₂, v₃] is given
by:
u x v = [u₂v₃ - u₃v₂, u₃v₁ - u₁v₃, u₁v₂ - u₂v₁]
For u = [2, -1, 3] and v = [-1, 3, 1], we have:
u x v = [(-1)(1) - (3)(3), (3)(2) - (2)(1), (2)(3) - (-1)(-1)]
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u x v = [-1 - 9, 6 - 2, 6 - 1]
u x v = [-10, 4, 5]
Therefore, the vector [-10, 4, 5] is perpendicular to both u = [2, -1, 3] and v
= [-1, 3, 1].
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AIOU Solved Assignment Code 1309 Mathematics III 2023 Assignment 2.pptx

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