# CLASSIFY CONIC SECTION DEFINED BY EQUATION

## Equation of Parabola

The variable which is not having square, then the parabola is symmetric about that particular axis.

## Equation of Ellipse

• If a2 is greater then, the ellipse will be symmetric about x-axis.
• If b2 is greater then, the ellipse will be symmetric about y-axis.

## Equation of Hyperbola

Equation of hyperbola which is symmetric about x-axis.

Equation of hyperbola which is symmetric about y-axis.

Classify each conic section.

Problem 1 :

x2 + y2 = 30

Solution:

x2 + y2 = 30

The given equation is in the form of x2 + y2 = r2

So, it is equation of circle. Center is (0, 0) and radius is √30.

Problem 2 :

x2 + y2 = 36

Solution:

The given equation is in the form of x2 + y2 = r2

r= 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 = y2

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)= x + 2

(y - k)= 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)= y - 3

(x - h)= 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

a2 = 1, b2 = 25

Since b2 is greater than a2, 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 :

-x2 + 10x + y - 21 = 0

Solution:

-x2 + 10x + y - 21 = 0

-(x2 - 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 :

-2y2 + x - 20y - 49 = 0

Solution:

-2y2 + x - 20y - 49 = 0

-2y2 - 20y + x = 49

-2(y2 + 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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