Conic Section	Standard Form	Key Formulas
Circle	(x – h)² + (y – k)² = r²	Center: (h, k) Radius: r
Ellipse	(x²/a²) + (y²/b²) = 1 (horizontal) (y²/a²) + (x²/b²) = 1 (vertical)	Center: (h, k) Vertices: (±a, 0) or (0, ±a) Co-vertices: (0, ±b) or (±b, 0) Foci: (±c, 0) or (0, ±c) c² = a² – b² Eccentricity: e = c/a
Parabola	y = a(x – h)² + k (vertical) x = a(y – k)² + h (horizontal)	Vertex: (h, k) Focus: (h, k + 1/(4a)) (vertical) Focus: (h + 1/(4a), k) (horizontal) Directrix: y = k – 1/(4a) (vertical) Directrix: x = h – 1/(4a) (horizontal)
Hyperbola	(x²/a²) – (y²/b²) = 1 (horizontal) (y²/a²) – (x²/b²) = 1 (vertical)	Center: (h, k) Vertices: (±a, 0) or (0, ±a) Co-vertices: (0, ±b) or (±b, 0) Foci: (±c, 0) or (0, ±c) c² = a² + b² Eccentricity: e = c/a Asymptotes: y = ±(b/a)x

Conic Sections Practice Worksheet

P1. Find the center and radius of the circle given by the equation: x² + y² + 4x – 6y + 4 = 0

P2. Determine the vertices, foci, and eccentricity of the ellipse: (x²/16) + (y²/9) = 1

P3. For the parabola y = 2x² – 4x + 5, find the vertex, axis of symmetry, and direction of opening.

P4. Identify the type of conic section represented by the equation: 4x² – 9y² = 36

P5. Find the equation of the circle with center (3, -2) and passing through the point (7, 1).

P6. Determine the coordinates of the foci for the hyperbola: (x²/25) – (y²/16) = 1

P7. Write the equation of a parabola with vertex at (2, -3) and focus at (2, 1).

P8. Find the eccentricity of the ellipse: 25x² + 9y² = 225

P9. Determine the equations of the asymptotes for the hyperbola: (y²/9) – (x²/16) = 1

P10. Given the general equation Ax² + By² + Cx + Dy + E = 0, what conditions on A and B determine whether this represents a circle?

Conic Sections Practice Problems

Problem 1: Identify the conic section: x² + y² = 25

Solution:

This is a circle with center (0, 0) and radius 5.

Problem 2: Find the center and vertices of the ellipse: (x – 3)²/16 + (y + 1)²/9 = 1

Solution:

Center: (3, -1)

Vertices: (3 ± 4, -1) = (7, -1) and (-1, -1)

Problem 3: Determine the vertex, focus, and directrix of the parabola: y = 2x² – 4x + 5

Solution:

Standard form: (x + 1)² = 2(y – 3)

Vertex: (-1, 3)

Focus: (-1, 3 + 1/4) = (-1, 3.25)

Directrix: y = 2.75

Problem 4: Find the center, vertices, and asymptotes of the hyperbola: (x + 2)²/25 – (y – 1)²/16 = 1

Solution:

Center: (-2, 1)

Vertices: (-2 ± 5, 1) = (3, 1) and (-7, 1)

Asymptotes: y – 1 = ±(4/5)(x + 2)

Problem 5: Identify the type of conic section: 4x² + 9y² – 24x – 54y + 81 = 0

Solution:

Rearranging: 4(x² – 6x) + 9(y² – 6y) = -81

Completing the square: 4(x² – 6x + 9) + 9(y² – 6y + 9) = 45

(x – 3)²/11.25 + (y – 3)²/5 = 1

This is an ellipse.

Problem 6: Find the eccentricity of the ellipse: 9x² + 16y² = 144

Solution:

Standard form: x²/16 + y²/9 = 1

a = 4, b = 3

e = √(1 – b²/a²) = √(1 – 9/16) = √(7/16) ≈ 0.661

Problem 7: Determine if the following points lie on the parabola y = x² – 2x + 3: (0, 3), (1, 2), (2, 3)

Solution:

For (0, 3): 3 = 0² – 2(0) + 3 = 3 (True)

For (1, 2): 2 = 1² – 2(1) + 3 = 2 (True)

For (2, 3): 3 = 2² – 2(2) + 3 = 3 (True)

All points lie on the parabola.

Problem 8: Find the latus rectum of the hyperbola: x²/16 – y²/9 = 1

Solution:

a = 4, b = 3

Latus rectum = 2b²/a

= 2(3²)/4 = 4.5

Problem 9: Determine the type of conic section and its properties: x² + 2y² + 4x – 8y + 4 = 0

Solution:

Rearranging: (x² + 4x) + 2(y² – 4y) = -4

Completing the square: (x + 2)² + 2(y – 2)² = 4

(x + 2)²/4 + (y – 2)²/2 = 1

This is an ellipse with center (-2, 2), a = 2, b = √2

Problem 10: Find the equation of the circle with center (3, -2) that passes through the point (7, 1)

Solution:

Using (x – h)² + (y – k)² = r²

r² = (7 – 3)² + (1 – (-2))² = 4² + 3² = 25

Equation: (x – 3)² + (y + 2)² = 25

Problem 11: Find the center and radius of the circle given by the equation: x² + y² – 6x + 8y – 11 = 0

Solution:

Step 1: Rearrange the equation to standard form (x – h)² + (y – k)² = r² (x² – 6x) + (y² + 8y) = 11 (x² – 6x + 9) + (y² + 8y + 16) = 11 + 9 + 16 (x – 3)² + (y + 4)² = 36

Step 2: Identify center and radius Center: (h, k) = (3, -4) Radius: r = √36 = 6

Therefore, the center is (3, -4) and the radius is 6.

Problem 12: Determine the vertices, foci, and eccentricity of the ellipse: (x²/25) + (y²/16) = 1

Solution:

Step 1: Identify a and b a² = 25, so a = 5 b² = 16, so b = 4

Step 2: Find vertices Vertices: (±a, 0) = (±5, 0)

Step 3: Calculate c c² = a² – b² = 25 – 16 = 9 c = 3

Step 4: Find foci Foci: (±c, 0) = (±3, 0)

Step 5: Calculate eccentricity e = c/a = 3/5 = 0.6

Therefore: Vertices: (5, 0) and (-5, 0) Foci: (3, 0) and (-3, 0) Eccentricity: 0.6

Problem 13: For the parabola y = 2x² – 4x + 5, find the vertex, axis of symmetry, and direction of opening.

Solution:

Step 1: Identify a, b, and c y = ax² + bx + c a = 2, b = -4, c = 5

Step 2: Find the x-coordinate of the vertex x = -b / (2a) = -(-4) / (2(2)) = 4/4 = 1

Step 3: Find the y-coordinate of the vertex y = 2(1)² – 4(1) + 5 = 2 – 4 + 5 = 3

Step 4: Determine the axis of symmetry The axis of symmetry is a vertical line through the vertex: x = 1

Step 5: Determine the direction of opening Since a > 0, the parabola opens upward.

Therefore: Vertex: (1, 3) Axis of symmetry: x = 1 Direction: Opens upward

Problem 14: Identify the center, vertices, and asymptotes of the hyperbola: (x²/16) – (y²/9) = 1

Solution:

Step 1: Identify a and b a² = 16, so a = 4 b² = 9, so b = 3

Step 2: Find the center The center is always at (0, 0) for this standard form.

Step 3: Find vertices Vertices: (±a, 0) = (±4, 0)

Step 4: Find equations of asymptotes y = ±(b/a)x y = ±(3/4)x

Therefore: Center: (0, 0) Vertices: (4, 0) and (-4, 0) Asymptotes: y = (3/4)x and y = -(3/4)x

Problem 15: Find the equation of the circle with center (-2, 3) and radius 5.

Solution:

Step 1: Use the standard form (x – h)² + (y – k)² = r² Where (h, k) is the center and r is the radius.

Step 2: Substitute the values (x – (-2))² + (y – 3)² = 5²

Step 3: Simplify (x + 2)² + (y – 3)² = 25

This is the equation of the circle.

Problem 16: Find the eccentricity of the ellipse: 4x² + 9y² = 36

Solution:

Step 1: Put the equation in standard form (x²/9) + (y²/4) = 1

Step 2: Identify a and b a² = 9, so a = 3 b² = 4, so b = 2

Step 3: Calculate c c² = a² – b² = 9 – 4 = 5 c = √5

Step 4: Calculate eccentricity e = c/a = √5/3 ≈ 0.745

Therefore, the eccentricity is √5/3 or approximately 0.745.

Frequently Asked Question (FAQs)

What are Four Types of Conic Sections?

The four types of conic section includes:

• Circles
• Ellipses
• Parabolas
• Hyperbolas

How is Eccentricity Related to Conic Sections?

For any conic section:

• Circle: e = 0
• Ellipse: 0 < e < 1
• Parabola: e = 1
• Hyperbola: e > 1

What is the Difference between Major and Minor Axes in an Ellipse?

Difference between Major and Minor axis for an ellipse is:

• Major axis is the longest diameter
• Minor axis is the shortest diameter

How many Focus Points does Each Conic Section Have?

For each conioc section:

• Circle: 1 (center)
• Ellipse: 2
• Parabola: 1
• Hyperbola: 2

What is the Standard Form Equation for a Circle?

Standard Form Equation for a Circle is (x-h)² + (y-k)² = r², where (h,k) is the center and r is the radius.