#### Solved Examples and Worksheet for Graphing Conic Sections

Q1Express the equation in vertex form and choose the graph.
y = - 4x2 - 16x - 7

A. y = - 4(x + 2)2 + 9; Graph 2
B. y = - 4(x + 2)2 - 9; Graph 4
C. y = - 4(x + 2)2 + 9; Graph 1
D. y = - 4(x + 2)2 + 9; Graph 3

Step: 1
y = - 4x2 - 16x - 7
Step: 2
y = - 4(x2 + 4x) - 7
[Take out the common factor.]
Step: 3
y = - 4(x2 + 4x + 4 - 4) - 7
[Add and subtract (42)2 = (2)2 = 4 inside the brackets.]
Step: 4
y = - 4(x2 + 4x + 4) + 16 - 7
[Rewrite the equation.]
Step: 5
y = - 4(x + 2)2 + 9
[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
Q2Express the equation in vertex form and identify the graph.
y = x2 + 4x + 1

A. y = (x - 2)2 - 3; Graph 3
B. y = (x + 2)2 - 3; Graph 2
C. y = (x + 2)2 - 3; Graph 4
D. y = (x + 2)2 - 3; Graph 1

Step: 1
y = x2 + 4x + 1
Step: 2
y = (x2 + 4x) + 1
Step: 3
y = (x2 + 4x + 4 - 4) + 1
[Add and subtract (42)2 = (2)2 = 4 inside the brackets.]
Step: 4
y = (x2 + 4x + 4) - 4 + 1
[Rewrite the equation.]
Step: 5
y = (x + 2)2 - 3
[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
Q3Express the equation in vertex form and choose the graph.
y = (2350)x2 - (235)x + 12

A. y = (2350)(x + 5)2 + (12); Graph 2
B. y = (2350)(x - 5)2 + (12); Graph 2
C. y = (2350)(x - 5)2 + (12); Graph 1
D. y = (2350)(x + 5)2 + (12); Graph 1

Step: 1
y = (2350)x2 - (235)x + 12
Step: 2
y = (2350)(x2 - 10x) + 12
[Take out the common factor.]
Step: 3
y = (2350)(x2 - 10x + 25 - 25) + 12
[Add and subtract (- 102)2 = (- 5)2 = 25 inside the brackets.]
Step: 4
y = (2350)(x2 -10x + 25) - (23×2550) + 12
[Rewrite the equation.]
Step: 5
y = (2350)(x2 - 10x + 25) - (232) + 12
[Simplify.]
Step: 6
y = (2350)(x - 5)2 + (12)
[Simplify.]
Step: 7
The vertex is (5, 12). 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 = (2350)(0 - 5)2 + (12) = (232) + (12) = 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 = (2350)(x - 5)2 + (12); Graph 2
Q4Identify the graph that represents the equation of the circle, x² + y² = 2.

A. Graph 4
B. Graph 1
C. Graph 2
D. Graph 3

Step: 1
x2 + y2 = 2
[Given.]
Step: 2
The general form of the equation of the circle is (x - h)2 + (y - k)2 = r2, 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
r2 = 2
Step: 6
r = 1.4
[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
Q5Identify the graph that represents the equation of the circle, x² + y² = 7.

A. Graph 4
B. Graph 1
C. Graph 2
D. Graph 3

Step: 1
x2 + y2 = 7
[Given.]
Step: 2
The general form of the equation of the circle is (x - h)2 + (y - k)2 = r2, 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
r2 = 7
Step: 6
r = 2.7
[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
Q6Identify the graph that represents the equation of the circle, x2 + y2 = 9.

A. Graph 2
B. Graph 3
C. Graph 4
D. Graph 1

Step: 1
x2 + y2 = 9
[Given.]
Step: 2
The general form of the equation of the circle is (x - h)2 + (y - k)2 = r2, 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
r2 = 9
Step: 6
r = 3
[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
Q7Identify the graph that represents the equation of the circle, x² + y² = 6.

A. Graph 4
B. Graph 1
C. Graph 3
D. Graph 2

Step: 1
x2 + y2 = 6
[Given.]
Step: 2
The general form of the equation of the circle is (x - h)2 + (y - k)2 = r2, 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
r2 = 6
Step: 6
r = 2.4
[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
Q8Identify the graph that represents the equation y210 + x25 = 1.

A. Graph 1
B. Graph 3
C. Graph 2
D. Graph 4

Step: 1
x210 + y25 = 1
[Given.]
Step: 2
The standard form of the equation of ellipse is (x - h)2a2 + (y - k)2b 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
a2 = 10
Step: 6
a = 10
[Take square root on both sides of the equation.]
Step: 7
b2 = 5
Step: 8
b = 5
[Take square root on both sides of the equation.]
Step: 9
Length of the major axis = 2a
= 210 units
Step: 10
Length of the minor axis = 2b
= 25 units
Step: 11
c2 = a2 - b2
[Relation between a, b, and c.]
Step: 12
c2 = 10 - 5 = 5
Step: 13
c = 5
[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 210 units minor axis as 25 units, x-intercepts as (± 10, 0), and y-intercepts as (± 5, 0).
Step: 16
Therefore, Graph 2 represents the given equation of the ellipse.
Correct Answer is :   Graph 2
Q9Identify the graph that represents the equation 4x² + 8y² = 32.

A. Graph 2
B. Graph 3
C. Graph 4
D. Graph 1

Step: 1
4x2 + 8y2 = 32
[Given.]
Step: 2
x28 + y24 = 1
[Divide both sides of the equation by 32.]
Step: 3
The standard form of the equation of ellipse is (x - h)2a2 + (y - k)2b 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
a2 = 8
Step: 7
a = 22
[Take square root on both sides of the equation.]
Step: 8
b2 = 4
Step: 9
b = 2
[Take square root on both sides of the equation.]
Step: 10
Length of the major axis = 2a
= 2(22) = 42 units
Step: 11
Length of the minor axis = 2b
= 2(2) = 4 units
Step: 12
c2 = a2 - b2
[Relation between a, b, and c.]
Step: 13
c2 = 8 - 4 = 4
Step: 14
c = 2
[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
Q10Identify the graph that represents the equation x² + 36y² = 36.

A. Graph 1
B. Graph 2
C. Graph 4
D. Graph 3

Step: 1
x2 + 36y2 = 36
[Given.]
Step: 2
x236 + y21 = 1
[Divide both sides of the equation by 36.]
Step: 3
The standard form of the equation of ellipse is (x - h)2a2 + (y - k)2b 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
a2 = 36
Step: 7
a = 6
[Take square root on both sides of the equation.]
Step: 8
b2 = 1
Step: 9
b = 1
[Take square root on both sides of the equation.]
Step: 10
Length of the major axis = 2a
= 2(6) = 12 units
Step: 11
Length of the minor axis = 2b
= 2(1) = 2 units
Step: 12
c2 = a2 - b2
[Relation between a, b, and c.]
Step: 13
c2 = 36 - 1 = 35
Step: 14
c = 35
[Take square root on both sides of the equation.]
Step: 15
So, the foci of the ellipse are (35, 0) and (- 35, 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 (35, 0) and (- 35, 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
Q11Identify the graph that represents the equation x² + 2y² = 2.

A. Graph 1
B. Graph 4
C. Graph 3
D. Graph 2

Step: 1
x2 + 2y2 = 2
[Given.]
Step: 2
x22 + y21 = 1
[Divide both sides of the equation by 2.]
Step: 3
The standard form of the equation of ellipse is (x - h)2a2 + (y - k)2b 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
a2 = 2
Step: 7
a = 2
[Take square root on both sides of the equation.]
Step: 8
b2 = 1
Step: 9
b = 1
[Take square root on both sides of the equation.]
Step: 10
Length of the major axis = 2a
= 22 units
Step: 11
Length of the minor axis = 2b
= 2(1) = 2 units
Step: 12
c2 = a2 - b2
[Relation between a, b, and c.]
Step: 13
c2 = 2 - 1 = 1
Step: 14
c = 1
[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
Q12Identify the graph that represents the equation x² - 2 = 2y².

A. Graph 4
B. Graph 2
C. Graph 3
D. Graph 1

Step: 1
The standard form of the equation x2a2 - y2b2 = 1.
Step: 2
Compare x² - 2 = 2y² to the standard form of the hyperbola equation.
Step: 3
The center of this hyperbola is at the origin. According to the equation, a² = 2, so a = ±2 and b² = 1, so b = ±1.
The coordinates of the vertices are (2, 0) and (- 2, 0).
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 x² - 2 = 2y².
Correct Answer is :   Graph 2
Q13Identify the graph that represents the equation x29 - y216 = 1.

A. Graph 4
B. Graph 1
C. Graph 2
D. Graph 3

Step: 1
The standard form of the equation x2a2 - y2b2 = 1.
Step: 2
Compare x29 - y216 = 1 to the standard form of the hyperbola equation.
Step: 3
The center of this hyperbola is at the origin. According to the equation, a² = 9, so a = ±3 and b² = 16, so b = ±4.
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 x29 - y216 = 1.
Correct Answer is :   Graph 3
Q14Identify the graph that represents the equation 3x² - 9y² = 27.

A. Graph 1
B. Graph 4
C. Graph 3
D. Graph 2

Step: 1
The standard form of the equation x2a2 - y2b2 = 1.
Step: 2
Compare 3x2 - 9y2 = 27 with the standard form of the hyperbola equation.
Step: 3
The center of this hyperbola is at the origin. According to the equation, a² = 9, so a = ±3 and b² = 3, so b = ±3.
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 3x2 - 9y2 = 27.
Correct Answer is :   Graph 1
Q15Identify the graph that represents the equation 25x² - 9y² = 225.

A. Graph 3
B. Graph 4
C. Graph 2
D. Graph 1

Step: 1
The standard form of the equation x2a2 - y2b2 = 1.
Step: 2
Compare 25x2 - 9y2 = 225 with the standard form of the hyperbola equation.
Step: 3
The center of this hyperbola is at the origin. According to the equation, a² = 9, so a = ±3 and b² = 25, so b = ±5.
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 25x2 - 9y2 = 225.
Correct Answer is :   Graph 2