Conic Sections: Class 11 Maths NCERT Guide

Understanding How Conic Sections are Formed

Imagine a double-napped cone, which looks like two ice-cream cones placed tip-to-tip. This cone has a central line called the axis. Now, imagine intersecting this cone with a flat sheet of paper (a plane). The shape of the curve formed at the intersection is a conic section.

The specific type of curve depends on the angle of the plane relative to the cone's axis. Let 'α' (alpha) be the angle the cone's slant edge makes with its axis, and 'β' (beta) be the angle the cutting plane makes with the axis.

1. Circle: If the plane is perpendicular to the axis (β = 90°), the intersection is a circle.
2. Ellipse: If the plane is tilted but still cuts through one cone (α < β < 90°), the curve is an elongated circle, called an ellipse.
3. Parabola: If the plane is tilted so that it is exactly parallel to the slant edge of the cone (β = α), the curve is a U-shaped parabola, which is open and does not close.
4. Hyperbola: If the plane is tilted even further (0 ≤ β < α), it cuts through both nappes of the cone, creating two separate, open curves. This pair of curves is a hyperbola.

Key Terminology of Conic Sections

Focus (F)

A fixed point (or two points for ellipse and hyperbola) used to define the conic section. The curve consists of points whose distances to the focus relate in a special way to another line or point.

Directrix

A fixed straight line used in the definition of a conic section. The shape of the conic is determined by the distances of its points from the focus and the directrix.

Eccentricity (e)

A constant ratio that defines the shape of a conic section. For any point P on the conic, the ratio of its distance from the focus (PF) to its perpendicular distance from the directrix is constant and is called eccentricity. For a Parabola, e = 1. For an Ellipse, 0 ≤ e < 1. For a Hyperbola, e > 1. For a circle, e = 0.

Latus Rectum

The chord through a focus perpendicular to the axis of the conic section. Its length is a parameter that describes the 'width' of the conic at its focus.

Worked Examples: Finding Properties from Equations

• Example 1: Parabola Find the focus, axis, equation of the directrix, and the length of the latus rectum for the parabola y² = 16x. Step 1: Identify the standard form. The given equation is of the form y² = 4ax. This represents a parabola opening to the right, with its vertex at the origin (0, 0) and axis along the x-axis. Step 2: Find the value of 'a'. Comparing y² = 16x with y² = 4ax, we get: 4a = 16 a = 16 / 4 = 4 Step 3: Determine the properties. Focus: The focus is at (a, 0). So, the focus is (4, 0). Axis: The axis of the parabola is the x-axis, so its equation is y = 0. Directrix: The equation of the directrix is x = -a. So, the directrix is x = -4. Latus Rectum: The length of the latus rectum is 4a. So, the length is 4(4) = 16. Final Answer: Focus: (4, 0), Axis: y=0, Directrix: x = -4, Length of Latus Rectum: 16.
• Example 2: Ellipse Find the coordinates of the foci, the vertices, the lengths of major and minor axes, and the eccentricity of the ellipse x²/25 + y²/9 = 1. Step 1: Identify the standard form and major axis. The equation is of the form x²/a² + y²/b² = 1. Since the denominator of x² (25) is greater than the denominator of y² (9), the major axis is along the x-axis. Step 2: Find the values of 'a', 'b', and 'c'. Comparing with the standard form, we have a² = 25 and b² = 9. So, a = 5 and b = 3. The relationship between a, b, and c for an ellipse is c² = a² - b². c² = 25 - 9 = 16 c = 4 Step 3: Determine the properties. Foci: The foci are at (±c, 0). So, the foci are at (±4, 0). Vertices: The vertices are at (±a, 0). So, the vertices are at (±5, 0). Major Axis: Length = 2a = 2(5) = 10. Minor Axis: Length = 2b = 2(3) = 6. * Eccentricity (e): e = c/a = 4/5. Final Answer: Foci: (±4, 0), Vertices: (±5, 0), Major Axis Length: 10, Minor Axis Length: 6, Eccentricity: 4/5.
• Example 3: Hyperbola Find the coordinates of the foci, the vertices, and the eccentricity of the hyperbola 9y² - 4x² = 36. Step 1: Convert to standard form. Divide the entire equation by 36 to make the right side equal to 1: (9y²/36) - (4x²/36) = 1 y²/4 - x²/9 = 1 Step 2: Identify the standard form and transverse axis. The equation is of the form y²/a² - x²/b² = 1. Since the y² term is positive, the transverse axis is along the y-axis. Step 3: Find the values of 'a', 'b', and 'c'. Comparing with the standard form, we have a² = 4 and b² = 9. So, a = 2 and b = 3. The relationship for a hyperbola is c² = a² + b². c² = 4 + 9 = 13 c = √13 Step 4: Determine the properties. Foci: The foci are on the y-axis at (0, ±c). So, the foci are at (0, ±√13). Vertices: The vertices are on the y-axis at (0, ±a). So, the vertices are at (0, ±2). * Eccentricity (e): e = c/a = √13 / 2. Final Answer: Foci: (0, ±√13), Vertices: (0, ±2), Eccentricity: √13 / 2.

Exam Traps and Key Identifiers

Pay close attention to the standard forms in your exams! A small sign change or swapping 'a' and 'b' can lead to a completely wrong answer.

• Identifying the Conic:
• Parabola: Only one variable is squared (e.g., y² = 4ax or x² = 4ay).
• Ellipse: Both variables are squared, have the same sign (both positive), and have different coefficients (e.g., x²/a² + y²/b² = 1).
• Hyperbola: Both variables are squared but have opposite signs (e.g., x²/a² - y²/b² = 1).
• Circle: A special ellipse where coefficients of x² and y² are equal (e.g., x² + y² = r²).

• Ellipse Trap: The value 'a²' is ALWAYS the larger denominator, regardless of whether it's under x² or y². This determines the major axis.
• Hyperbola Trap: The value 'a²' is ALWAYS under the positive term. This determines the transverse axis (the axis that the hyperbola intersects).

Practice Questions with Solutions

• Q: Find the equation of the parabola with focus at (0, -3) and directrix y = 3. A: Step 1: Identify the type and orientation. The focus (0, -3) is on the y-axis, and the directrix y=3 is horizontal. This means the parabola opens downwards and its vertex is at the origin (0,0), halfway between the focus and directrix. The standard equation is x² = -4ay. Step 2: Determine the value of 'a'. The distance from the vertex (0,0) to the focus (0,-3) is 'a'. So, a = 3. Step 3: Substitute 'a' into the standard equation. x² = -4(3)y Final answer: The equation of the parabola is x² = -12y.
• Q: Find the equation of the ellipse whose vertices are at (±5, 0) and foci are at (±4, 0). A: Step 1: Determine the major axis and values of 'a' and 'c'. The vertices and foci are on the x-axis, so the major axis is along the x-axis. The standard form is x²/a² + y²/b² = 1. From the vertices (±a, 0) = (±5, 0), we get a = 5. From the foci (±c, 0) = (±4, 0), we get c = 4. Step 2: Find the value of 'b'. For an ellipse, c² = a² - b². So, b² = a² - c². b² = 5² - 4² = 25 - 16 = 9. Step 3: Write the final equation. Substitute a² = 25 and b² = 9 into the standard form. Final answer: The equation of the ellipse is x²/25 + y²/9 = 1.
• Q: Identify the conic section represented by the equation x² + y² - 4x + 6y - 3 = 0 and find its center and radius. A: Step 1: Rearrange and complete the square for x and y terms. Group x terms and y terms: (x² - 4x) + (y² + 6y) = 3. To complete the square for x, add (4/2)² = 4. To complete the square for y, add (6/2)² = 9. Add these to both sides: (x² - 4x + 4) + (y² + 6y + 9) = 3 + 4 + 9. Step 2: Rewrite in standard circle form. (x - 2)² + (y + 3)² = 16. This is the equation of a circle in the form (x - h)² + (y - k)² = r². Step 3: Identify the center and radius. By comparing, the center (h, k) is (2, -3) and the radius squared r² is 16, so the radius r is 4. Final answer: The conic is a circle with center (2, -3) and radius 4.
• Q: Find the equation for the hyperbola with vertices at (0, ±6) and eccentricity e = 5/3. A: Step 1: Identify the transverse axis and the value of 'a'. Since the vertices (0, ±a) are on the y-axis, the transverse axis is vertical. The standard form is y²/a² - x²/b² = 1. From the vertices, we have a = 6. Step 2: Use the eccentricity to find 'c'. We know e = c/a. So, c = e a = (5/3) 6 = 10. Step 3: Find the value of 'b'. For a hyperbola, c² = a² + b². So, b² = c² - a². b² = 10² - 6² = 100 - 36 = 64. Step 4: Write the final equation. Substitute a² = 36 and b² = 64 into the standard form. Final answer: The equation of the hyperbola is y²/36 - x²/64 = 1.

Frequently Asked Questions