The variable which is not having square, then the parabola is symmetric about that particular axis.
Equation of hyperbola which is symmetric about x-axis.
Equation of hyperbola which is symmetric about y-axis.
Classify each conic section.
Problem 1 :
x^{2} + y^{2} = 30
Solution:
x^{2} + y^{2} = 30
The given equation is in the form of x^{2} + y^{2} = r^{2}
So, it is equation of circle. Center is (0, 0) and radius is √30.
Problem 2 :
x^{2} + y^{2} = 36
Solution:
The given equation is in the form of x^{2} + y^{2} = r^{2}
r^{2 }= 36
r = 6
So, it is equation of circle. Center is (0, 0) and radius is 6
Problem 3 :
Solution:
In the equation is defined in term of x and y. Both variables are having square.
It is the equation of ellipse and it is symmetric about y-axis.
Problem 4 :
x = y^{2}
Solution:
The equation is defined in terms of x and y. One of the variable is having square the other variable is not having square.
So, the given equation is parabola.
Vertex of the parabola : V(0, 0)
Axis of symmetry : Symmetric about x-axis
Problem 5 :
x = (y + 4)^{2} - 2
Solution:
x = (y + 4)^{2} - 2
The equation is defined in terms of x and y. One of the variable is having square the other variable is not having square.
So, the given equation is parabola.
Converting into standard form :
x + 2 = (y + 4)^{2}
(y + 4)^{2 }= x + 2
(y - k)^{2 }= x - h
(h, k) ==> (-2, -4)
Vertex of the parabola : V(-2, -4)
Axis of symmetry : Symmetric about x-axis and x = -2
Problem 6 :
Solution:
In the equation is defined in term of and y. There is a minus between 2 variables. Both variables are having square. Equation equal to 1.
Hence, it is hyperbola.
Problem 7 :
y = (x - 1)^{2} + 3
Solution:
y = (x - 1)^{2} + 3
Converting into standard form :
y - 3 = (x - 1)^{2}
(x - 1)^{2 }= y - 3
(x - h)^{2 }= 4a(y - k)
h = 1 and k = 3
The parabola is symmetric about y-axis and it is open upward.
Vertex of the parabola : V(-2, -4)
Axis of symmetry : Symmetric about y-axis and y = 1
Problem 8 :
Solution:
In the equation both variables x and y are having square. It exactly matches with the general form
a^{2} = 1, b^{2} = 25
Since b^{2} is greater than a^{2}, it must be symmetric about y-axis.
Center : (h, k) ==> (1, 0)
Classify each conic section and write its equation in standard form.
Problem 9 :
-x^{2} + 10x + y - 21 = 0
Solution:
-x^{2} + 10x + y - 21 = 0
-(x^{2} - 10x + 25) + y = 21 - 25
-(x - 5)^{2} + y = -4
y = (x - 5)^{2} - 4
(x - 5)^{2 }= y + 4
(x - h)^{2 }= 4a(y - k)
h = 5, k = -4
Vertex : (h, k) ==> (5, -4)
Axis of symmetry : The parabola is symmetric about y-axis
Problem 10 :
-2y^{2} + x - 20y - 49 = 0
Solution:
-2y^{2} + x - 20y - 49 = 0
-2y^{2} - 20y + x = 49
-2(y^{2} + 10y + 25) + x = 49 - 50
-2(y + 5)^{2} + x = -1
x = 2(y + 5)^{2} - 1
x + 1 = 2(y + 5)^{2}
(y + 5)^{2 }= (1/2)(x + 1)
(y - k)^{2 }= 4a(x - h)
h = -1, k = -5
Vertex : (h, k) ==> (-1, -5)
Axis of symmetry : The parabola is symmetric about x-axis
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May 21, 24 08:51 AM
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