Step: 1

Step: 2

[Take out the common factor.]

Step: 3

[Add and subtract (4 2 )^{2} = (2)^{2} = 4 inside the brackets.]

Step: 4

[Rewrite the equation.]

Step: 5

[Simplify.]

Step: 6

The vertex is (- 2, 9).
The axis of symmetry is the line with the equation x = - 2.

Step: 7

Plot the vertex and draw the axis of symmetry.

Step: 8

Find another point.

Step: 9

When x = 0,
y = - 4(0 + 2)^{2} + 9
= - 16 + 9 = - 7.
The point is (0, - 7).

Step: 10

The reflection of (0, - 7) is (- 4, - 7).
Plot it.

Step: 11

Connect the points with a smooth curve.

Correct Answer is : y = - 4(x + 2)^{2} + 9; Graph 1

Step: 1

Step: 2

Step: 3

[Add and subtract (4 2 )^{2} = (2)^{2} = 4 inside the brackets.]

Step: 4

[Rewrite the equation.]

Step: 5

[Simplify.]

Step: 6

The vertex is (- 2, - 3).
The axis of symmetry is the line with the equation x = - 2.

Step: 7

Plot the vertex and draw the axis of symmetry.

Step: 8

Find another point.

Step: 9

When x = 0,
y = (0 + 2)^{2} - 3
= 4 - 3 = 1
The point is (0, 1).

Step: 10

The reflection of (0, 1) is (- 4, 1).
Plot it.

Step: 11

Connect the points with a smooth curve.

Correct Answer is : y = (x + 2)^{2} - 3; Graph 4

Step: 1

Step: 2

[Take out the common factor.]

Step: 3

[Add and subtract (- 10 2 )^{2} = (- 5)^{2} = 25 inside the brackets.]

Step: 4

[Rewrite the equation.]

Step: 5

[Simplify.]

Step: 6

[Simplify.]

Step: 7

The vertex is (5, 1 2 ).
The axis of symmetry is the line with the equation x = 5.

Step: 8

Plot the vertex and draw the axis of symmetry.

Step: 9

Find another point.

Step: 10

When x = 0,
y = (23 50 )(0 - 5)^{2} + (1 2 )
= (23 2 ) + (1 2 ) = 12
The point is (0, 12).

Step: 11

The reflection of (0, 12) is (10, 12).
Plot it.

Step: 12

Connect the points with a smooth curve.

Correct Answer is : y = (23 50 )(x - 5)^{2} + (1 2 ); Graph 2

Step: 1

[Given.]

Step: 2

The general form of the equation of the circle is (x - h )^{2} + (y - k )^{2} = r ^{2}, where (x , y ) are the coordinates
of a point on the circle, (h , k ) is the centre of the circle, and r is the radius
of the circle.

Step: 3

Compare the given equation with the general form of the circle equation.

Step: 4

(h , k ) = (0, 0)

Step: 5

Step: 6

[Take square root on both sides of the equation.]

Step: 7

From the given graphs, we can observe that the circle in Graph 1 has centre as origin and radius as 1.4.

Step: 8

Therefore, Graph 1 represents the given equation of the circle.

Correct Answer is : Graph 1

Step: 1

[Given.]

Step: 2

The general form of the equation of the circle is (x - h )^{2} + (y - k )^{2} = r ^{2}, where (x , y ) are the coordinates
of a point on the circle, (h , k ) is the centre of the circle, and r is the radius
of the circle.

Step: 3

Compare the given equation with the general form of the circle equation.

Step: 4

(h , k ) = (0, 0)

Step: 5

Step: 6

[Take square root on both sides of the equation.]

Step: 7

From the given graphs, we can observe that the circle in Graph 2 has centre as origin and radius as 2.7.

Step: 8

Therefore, Graph 2 represents the given equation of the circle.

Correct Answer is : Graph 2

Step: 1

[Given.]

Step: 2

The general form of the equation of the circle is (x - h )^{2} + (y - k )^{2} = r ^{2}, where (x , y ) are the coordinates
of a point on the circle, (h , k ) is the centre of the circle, and r is the radius
of the circle.

Step: 3

Compare the given equation with the general form of the circle equation.

Step: 4

(h , k ) = (0, 0)

Step: 5

Step: 6

[Take square root on both sides of the equation.]

Step: 7

From the given graphs, we can observe that the circle in Graph 1 has centre as origin and radius as 3.

Step: 8

Therefore, Graph 1 represents the given equation of the circle.

Correct Answer is : Graph 1

Step: 1

[Given.]

Step: 2

Step: 3

Compare the given equation with the general form of the circle equation.

Step: 4

(h , k ) = (0, 0)

Step: 5

Step: 6

[Take square root on both sides of the equation.]

Step: 7

From the given graphs, we can observe that the circle in Graph 2 has centre as origin and radius as 2.4.

Step: 8

Therefore, Graph 2 represents the given equation of the circle.

Correct Answer is : Graph 2

Step: 1

[Given.]

Step: 2

The standard form of the equation of ellipse is ( x - h ) 2 a 2 + ( y - k ) 2 b 2 = 1.

[(x , y ) are the coordinates of a point on the ellipse and (h , k ) is the center of the ellipse.]

Step: 3

Compare the given equation with the standard form of the equation of ellipse.

Step: 4

The center of the ellipse is at (0, 0).

Step: 5

Step: 6

[Take square root on both sides of the equation.]

Step: 7

Step: 8

[Take square root on both sides of the equation.]

Step: 9

Length of the major axis = 2a

= 21 0 units

= 2

Step: 10

Length of the minor axis = 2b

= 25 units

= 2

Step: 11

[Relation between a , b , and c .]

Step: 12

Step: 13

[Take square root on both sides of the equation.]

Step: 14

So, the foci of the ellipse are (5 , 0) and (- 5 , 0).

[The foci of the ellipse are (c , 0) and (- c , 0).]

Step: 15

From the given graphs, we can observe that the ellipse in Graph 2 have center at the origin, foci as (5 , 0) and (- 5 , 0), length of the major axis as 21 0 units minor axis as 25 units, x -intercepts as (± 1 0 , 0), and y -intercepts as (± 5 , 0).

Step: 16

Therefore, Graph 2 represents the given equation of the ellipse.

Correct Answer is : Graph 2

Step: 1

4x 2 + 8y 2 = 32

[Given.]

Step: 2

[Divide both sides of the equation by 32.]

Step: 3

The standard form of the equation of ellipse is ( x - h ) 2 a 2 + ( y - k ) 2 b 2 = 1.

[(x , y ) are the coordinates of a point on the ellipse and (h , k ) is the center of the ellipse.]

Step: 4

Compare the given equation with the standard form of the equation of ellipse.

Step: 5

The center of the ellipse is at (0, 0).

Step: 6

Step: 7

[Take square root on both sides of the equation.]

Step: 8

Step: 9

[Take square root on both sides of the equation.]

Step: 10

Length of the major axis = 2a

= 2(22 ) = 42 units

= 2(2

Step: 11

Length of the minor axis = 2b

= 2(2) = 4 units

= 2(2) = 4 units

Step: 12

[Relation between a , b , and c .]

Step: 13

Step: 14

[Take square root on both sides of the equation.]

Step: 15

So, the foci of the ellipse are (2, 0) and (- 2, 0).

[The foci of the ellipse are (c , 0) and (- c , 0).]

Step: 16

From the given graphs, we can observe that the ellipse in Graph 4 have center at the origin, foci as (2, 0) and (- 2, 0), length of the major axis as 42 units and minor axis as 4 units, x -intercepts as (± 22 , 0), and y -intercepts as (± 2, 0).

Step: 17

Therefore, Graph 4 represents the given equation of the ellipse.

Correct Answer is : Graph 4

Step: 1

[Given.]

Step: 2

[Divide both sides of the equation by 36.]

Step: 3

The standard form of the equation of ellipse is ( x - h ) 2 a 2 + ( y - k ) 2 b 2 = 1.

[(x , y ) are the coordinates of a point on the ellipse and (h , k ) is the center of the ellipse.]

Step: 4

Compare the given equation with the standard form of the equation of ellipse.

Step: 5

The center of the ellipse is at (0, 0).

Step: 6

Step: 7

[Take square root on both sides of the equation.]

Step: 8

Step: 9

[Take square root on both sides of the equation.]

Step: 10

Length of the major axis = 2a

= 2(6) = 12 units

= 2(6) = 12 units

Step: 11

Length of the minor axis = 2b

= 2(1) = 2 units

= 2(1) = 2 units

Step: 12

[Relation between a , b , and c .]

Step: 13

Step: 14

[Take square root on both sides of the equation.]

Step: 15

So, the foci of the ellipse are (3 5 , 0) and (- 3 5 , 0).

[The foci of the ellipse are (c , 0) and (- c , 0).]

Step: 16

From the given graphs, we can observe that the ellipse in Graph 3 have center at the origin, foci as (3 5 , 0) and (- 3 5 , 0), length of the major axis as 12 units and minor axis as 2 units, x -intercepts as (± 6, 0), and y -intercepts as (± 1, 0).

Step: 17

Therefore, Graph 3 represents the given equation of the ellipse.

Correct Answer is : Graph 3

Step: 1

[Given.]

Step: 2

[Divide both sides of the equation by 2.]

Step: 3

The standard form of the equation of ellipse is ( x - h ) 2 a 2 + ( y - k ) 2 b 2 = 1.

[(x , y ) are the coordinates of a point on the ellipse and (h , k ) is the center of the ellipse.

Step: 4

Compare the given equation with the standard form of the equation of ellipse.

Step: 5

The center of the ellipse is at (0, 0).

Step: 6

Step: 7

[Take square root on both sides of the equation.]

Step: 8

Step: 9

[Take square root on both sides of the equation.]

Step: 10

Length of the major axis = 2a

= 22 units

= 2

Step: 11

Length of the minor axis = 2b

= 2(1) = 2 units

= 2(1) = 2 units

Step: 12

[Relation between a , b , and c .]

Step: 13

Step: 14

[Take square root on both sides of the equation.]

Step: 15

So, the foci of the ellipse are (1, 0) and (- 1, 0).

[The foci of the ellipse are (c , 0) and (- c , 0).]

Step: 16

From the given graphs, we can observe that the ellipse in Graph 2 have center at the origin, foci as (1, 0) and (- 1, 0), length of the major axis as 2rout(2) units and minor axis as 2 units, x -intercepts as (± 2 , 0), and y -intercepts as (± 1, 0).

Step: 17

Therefore, Graph 2 represents the given equation of the ellipse.

Correct Answer is : Graph 2

Step: 1

The standard form of the equation x 2 a 2 - y 2 b 2 = 1.

Step: 2

Compare x ² - 2 = 2y ² to the standard form of the hyperbola equation.

Step: 3

The center of this hyperbola is at the origin. According to the equation, a ² = 2, so a = ±2 and b ² = 1, so b = ±1.

The coordinates of the vertices are (2 , 0) and (- 2 , 0).

The coordinates of the vertices are (

Step: 4

So, the hyperbola passes through the points (2 , 0) and (- 2 , 0).

Step: 5

The points (0, - 1) and (0, 1) and the points (2 , 0) and (- 2 , 0) can be used to form a rectangle whose diagonals are the asymptotes of the hyperbola.

Step: 6

Now we can sketch the hyperbola. The graph goes through the points (2 , 0) and (- 2 , 0) and get closer and closer to the asymptotes as we get farther and farther from the center of the figure.

Step: 7

Therefore, Graph 2 represents the equation x ² - 2 = 2y ².

Correct Answer is : Graph 2

Step: 1

The standard form of the equation x 2 a 2 - y 2 b 2 = 1.

Step: 2

Compare x 2 9 - y 2 1 6 = 1 to the standard form of the hyperbola equation.

Step: 3

The center of this hyperbola is at the origin. According to the equation, a ² = 9, so a = ±3 and b ² = 16, so b = ±4.

The coordinates of the vertices are (3, 0) and (- 3, 0).

The coordinates of the vertices are (3, 0) and (- 3, 0).

Step: 4

So, the hyperbola passes through the points (3, 0) and (- 3, 0).

Step: 5

The points (0, - 4) and (0, 4) and the points (3, 0) and (- 3, 0) can be used to form a rectangle whose diagonals are the asymptotes of the hyperbola.

Step: 6

Now we can sketch the hyperbola. The graph goes through the points (3, 0) and (- 3, 0) and get closer and closer to the asymptotes as we get farther and farther from the center of the figure.

Step: 7

Therefore, Graph 3 represents the equation x 2 9 - y 2 1 6 = 1.

Correct Answer is : Graph 3

Step: 1

The standard form of the equation x 2 a 2 - y 2 b 2 = 1.

Step: 2

Compare 3x 2 - 9y 2 = 27 with the standard form of the hyperbola equation.

Step: 3

The center of this hyperbola is at the origin. According to the equation, a ² = 9, so a = ±3 and b ² = 3, so b = ±3 .

The coordinates of the vertices are (3, 0) and (- 3, 0).

The coordinates of the vertices are (3, 0) and (- 3, 0).

Step: 4

So, the hyperbola passes through the points (3, 0) and (- 3, 0).

Step: 5

The points (0, - 3 ) and (0, 3 ) and the points (3, 0) and (- 3, 0) can be used to form a rectangle whose diagonals are the asymptotes of the hyperbola.

Step: 6

Now we can sketch the hyperbola. The graph goes through the points (3, 0) and (- 3, 0) and get closer and closer to the asymptotes as we get farther and farther from the center of the figure.

Step: 7

Therefore, Graph 1 represents the equation 3x 2 - 9y 2 = 27.

Correct Answer is : Graph 1

Step: 1

The standard form of the equation x 2 a 2 - y 2 b 2 = 1.

Step: 2

Compare 25x 2 - 9y 2 = 225 with the standard form of the hyperbola equation.

Step: 3

The center of this hyperbola is at the origin. According to the equation, a ² = 9, so a = ±3 and b ² = 25, so b = ±5.

The coordinates of the vertices are (3, 0) and (- 3, 0).

The coordinates of the vertices are (3, 0) and (- 3, 0).

Step: 4

So, the hyperbola passes through the points (3, 0) and (- 3, 0).

Step: 5

The points (0, - 5) and (0, 5) and the points (3, 0) and (- 3, 0) can be used to form a rectangle whose diagonals are the asymptotes of the hyperbola.

Step: 6

Now we can sketch the hyperbola. The graph goes through the points (3, 0) and (- 3, 0) and get closer and closer to the asymptotes as we get farther and farther from the center of the figure.

Step: 7

Therefore, Graph 2 represents the equation 25x 2 - 9y 2 = 225.

Correct Answer is : Graph 2

- Identifying and Using the Slopes of Parallel and Perpendicular Lines-Geometry-Solved Examples
- Writing Linear Equations of Parallel and Perpendicular Lines-Geometry-Solved Examples
- Equations of Circles-Geometry-Solved Examples
- Standard Forms and Equations of Ellipses and Parabolas-Geometry-Solved Examples
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- Conic Section