Conic Sections Ex 11.3 Class 11 Maths NCERT Solutions

Welcome, Class 11 students, to a crucial exploration of Conic Sections, specifically focusing on Exercise 11.3 from your NCERT textbook! This exercise dives deep into the fascinating worlds of Ellipses and Hyperbolas, two fundamental shapes derived from slicing a double-napped cone. Understanding these figures isn't just about solving textbook problems; it's about appreciating the mathematics behind orbits of planets, designs of arch bridges, and even the acoustics of concert halls.

In this comprehensive guide, we'll break down the definitions, standard equations, and key properties of ellipses and hyperbolas. You'll learn to identify their foci, vertices, axes, eccentricity, and latus rectum, and confidently solve problems related to finding their equations or characteristics. By the end of this session, you'll be well-equipped to tackle any question from Conic Sections Exercise 11.3, enhancing your conceptual clarity and problem-solving skills for your CBSE exams and beyond.

Understanding the Ellipse

An ellipse is a closed curve defined as the set of all points in a plane such that the sum of their distances from two fixed points (called foci, F1 and F2) is constant. This constant sum is equal to the length of the major axis, denoted as 2a.

There are two standard forms for the equation of an ellipse centered at the origin (0,0):

  1. Horizontal Ellipse (Major axis along X-axis):

x^2/a^2 + y^2/b^2 = 1, where a > b.

  • Vertices: (±a, 0)
  • Foci: (±c, 0)
  • Major Axis Length: 2a
  • Minor Axis Length: 2b
  • Relationship: c^2 = a^2 - b^2 or a^2 = b^2 + c^2 (Note: a is always the semi-major axis length, and b is the semi-minor axis length. 'a' is associated with the larger denominator).
  • Eccentricity (e): e = c/a, where 0 < e < 1.
  • Length of Latus Rectum: 2b^2/a
  1. Vertical Ellipse (Major axis along Y-axis):

x^2/b^2 + y^2/a^2 = 1, where a > b.

  • Vertices: (0, ±a)
  • Foci: (0, ±c)
  • Major Axis Length: 2a
  • Minor Axis Length: 2b
  • Relationship: c^2 = a^2 - b^2 or a^2 = b^2 + c^2
  • Eccentricity (e): e = c/a, where 0 < e < 1.
  • Length of Latus Rectum: 2b^2/a

The key to solving problems is correctly identifying whether the major axis is horizontal or vertical, which depends on whether a^2 is under x^2 or y^2 respectively, where a is the larger of the two denominators.

Understanding the Hyperbola

A hyperbola is an open curve defined as the set of all points in a plane such that the absolute difference of their distances from two fixed points (called foci, F1 and F2) is constant. This constant difference is equal to the length of the transverse axis, denoted as 2a.

Like ellipses, hyperbolas also have two standard forms centered at the origin (0,0):

  1. Horizontal Hyperbola (Transverse axis along X-axis):

x^2/a^2 - y^2/b^2 = 1

  • Vertices: (±a, 0)
  • Foci: (±c, 0)
  • Transverse Axis Length: 2a
  • Conjugate Axis Length: 2b
  • Relationship: c^2 = a^2 + b^2 (Note: For hyperbolas, a^2 is always under the positive term).
  • Eccentricity (e): e = c/a, where e > 1.
  • Length of Latus Rectum: 2b^2/a
  • Asymptotes: y = ±(b/a)x
  1. Vertical Hyperbola (Transverse axis along Y-axis):

y^2/a^2 - x^2/b^2 = 1

  • Vertices: (0, ±a)
  • Foci: (0, ±c)
  • Transverse Axis Length: 2a
  • Conjugate Axis Length: 2b
  • Relationship: c^2 = a^2 + b^2
  • Eccentricity (e): e = c/a, where e > 1.
  • Length of Latus Rectum: 2b^2/a
  • Asymptotes: y = ±(a/b)x

For hyperbolas, 'a' is always the distance from the center to a vertex, and a^2 is the denominator of the positive term. 'b' is the distance from the center to a co-vertex.

Worked Examples from Ex 11.3

  • Example 1: Find the equation of the ellipse satisfying the given conditions: Vertices (±5, 0), Foci (±4, 0). Step 1: Identify the type of ellipse. Since the vertices and foci are on the x-axis, it is a horizontal ellipse. Its standard equation is x^2/a^2 + y^2/b^2 = 1. Step 2: Determine 'a' and 'c'. From the vertices (±a, 0), we get a = 5. From the foci (±c, 0), we get c = 4. Step 3: Calculate 'b' using the relationship c^2 = a^2 - b^2. 4^2 = 5^2 - b^2 16 = 25 - b^2 b^2 = 25 - 16 b^2 = 9 Step 4: Substitute a^2 and b^2 into the standard equation. x^2/5^2 + y^2/9 = 1 x^2/25 + y^2/9 = 1 Final answer: The equation of the ellipse is x^2/25 + y^2/9 = 1.
  • Example 2: Find the foci, vertices, eccentricity, and the length of the latus rectum of the hyperbola y^2/9 - x^2/16 = 1. Step 1: Identify the standard form. The given equation y^2/9 - x^2/16 = 1 matches the vertical hyperbola form y^2/a^2 - x^2/b^2 = 1. Step 2: Determine 'a^2' and 'b^2'. From y^2/a^2 = y^2/9, we get a^2 = 9, so a = 3. From x^2/b^2 = x^2/16, we get b^2 = 16, so b = 4. Step 3: Calculate 'c' using the relationship c^2 = a^2 + b^2. c^2 = 9 + 16 c^2 = 25 c = 5 Step 4: Find the foci and vertices. For a vertical hyperbola, Foci are (0, ±c) = (0, ±5). For a vertical hyperbola, Vertices are (0, ±a) = (0, ±3). Step 5: Calculate eccentricity (e). e = c/a = 5/3. Step 6: Calculate the length of the latus rectum. Length of Latus Rectum = 2b^2/a = 2(16)/3 = 32/3. Final answer: Foci (0, ±5), Vertices (0, ±3), Eccentricity 5/3, Latus Rectum 32/3.
  • Example 3: Find the equation of the hyperbola with Foci (0, ±13) and the length of the transverse axis is 24. Step 1: Identify the type of hyperbola. Since the foci are on the y-axis, it is a vertical hyperbola. Its standard equation is y^2/a^2 - x^2/b^2 = 1. Step 2: Determine 'c' and 'a'. From Foci (0, ±c), we get c = 13. Length of transverse axis = 2a = 24, so a = 12. Step 3: Calculate 'b' using the relationship c^2 = a^2 + b^2. 13^2 = 12^2 + b^2 169 = 144 + b^2 b^2 = 169 - 144 b^2 = 25 Step 4: Substitute a^2 and b^2 into the standard equation. y^2/12^2 - x^2/25 = 1 y^2/144 - x^2/25 = 1 Final answer: The equation of the hyperbola is y^2/144 - x^2/25 = 1.

Exam Tips for Conic Sections Ex 11.3

Mastering Exercise 11.3 requires careful attention to detail. Here are some crucial tips to avoid common mistakes and score well:

  1. Identify the Conic Section First: Always check if it's an ellipse or a hyperbola. For x^2/A + y^2/B = 1, if A and B have the same sign, it's an ellipse. If they have opposite signs, it's a hyperbola. For Ax^2 + By^2 = C, if A and B have same sign, it's ellipse; if opposite, hyperbola.
  2. Determine Orientation (Horizontal/Vertical):
  • Ellipse: The major axis is along the axis corresponding to the larger denominator. If a^2 is under x^2, it's horizontal. If a^2 is under y^2, it's vertical. Remember a > b for ellipses.
  • Hyperbola: The transverse axis is along the axis corresponding to the positive term. If x^2/a^2 - y^2/b^2 = 1, it's horizontal. If y^2/a^2 - x^2/b^2 = 1, it's vertical. Here, a^2 is always the denominator of the positive term.
  1. Correct Relationship between a, b, c:
  • Ellipse: c^2 = a^2 - b^2 (foci are inside the ellipse, so c < a)
  • Hyperbola: c^2 = a^2 + b^2 (foci are outside the vertices, so c > a)
  • Memorizing these correctly is vital.
  1. Practice Drawing Diagrams: Even a rough sketch can help visualize the given information (foci, vertices) and confirm the orientation of the conic section.
  2. Be Careful with Signs: A single sign error, especially in the hyperbola equation or the c^2 relationship, can lead to entirely wrong results. Double-check your calculations.

Practice Questions with Solutions

  • Q: Find the equation of the ellipse whose major axis is along the x-axis, and passes through the points (4, 3) and (6, 2). A: Step 1: Assume the equation of the ellipse is x^2/a^2 + y^2/b^2 = 1 as the major axis is along the x-axis. Step 2: Substitute the given points into the equation. For (4, 3): 16/a^2 + 9/b^2 = 1 (Equation 1) For (6, 2): 36/a^2 + 4/b^2 = 1 (Equation 2) Step 3: Solve the system of linear equations for 1/a^2 and 1/b^2. Let X = 1/a^2 and Y = 1/b^2. 16X + 9Y = 1 36X + 4Y = 1 Multiply the first equation by 4 and the second by 9: 64X + 36Y = 4 324X + 36Y = 9 Subtract the first new equation from the second new equation: (324X - 64X) + (36Y - 36Y) = 9 - 4 260X = 5 X = 5/260 = 1/52 So, 1/a^2 = 1/52, which means a^2 = 52. Step 4: Substitute X = 1/52 into 16X + 9Y = 1. 16(1/52) + 9Y = 1 4/13 + 9Y = 1 9Y = 1 - 4/13 9Y = 9/13 Y = 1/13 So, 1/b^2 = 1/13, which means b^2 = 13. Step 5: Substitute a^2 and b^2 into the ellipse equation. x^2/52 + y^2/13 = 1 Final answer: The equation of the ellipse is x^2/52 + y^2/13 = 1.
  • Q: Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 9y^2 - 4x^2 = 36. A: Step 1: Convert the given equation to the standard form of a hyperbola. Divide the entire equation by 36. 9y^2/36 - 4x^2/36 = 36/36 y^2/4 - x^2/9 = 1 Step 2: Identify a^2 and b^2. This is a vertical hyperbola because the y^2 term is positive. a^2 = 4, so a = 2. b^2 = 9, so b = 3. Step 3: Calculate c using c^2 = a^2 + b^2. c^2 = 4 + 9 = 13 c = sqrt(13). Step 4: Determine foci, vertices, eccentricity, and latus rectum. Foci: For a vertical hyperbola, foci are (0, ±c) = (0, ±sqrt(13)). Vertices: For a vertical hyperbola, vertices are (0, ±a) = (0, ±2). Eccentricity (e): e = c/a = sqrt(13)/2. Length of latus rectum: 2b^2/a = 2(9)/2 = 9. Final answer: Foci (0, ±sqrt(13)), Vertices (0, ±2), Eccentricity sqrt(13)/2, Length of Latus Rectum 9.
  • Q: Find the equation of the hyperbola with vertices (±7, 0) and eccentricity e = 4/3. A: Step 1: Identify the type of hyperbola. Since vertices are (±7, 0), the transverse axis is along the x-axis. It is a horizontal hyperbola with the standard equation x^2/a^2 - y^2/b^2 = 1. Step 2: Determine 'a' and 'c'. From vertices (±a, 0), a = 7. Given eccentricity e = c/a = 4/3. Substitute a = 7: c/7 = 4/3 c = 28/3. Step 3: Calculate 'b' using the relationship c^2 = a^2 + b^2. (28/3)^2 = 7^2 + b^2 784/9 = 49 + b^2 b^2 = 784/9 - 49 b^2 = 784/9 - 441/9 b^2 = (784 - 441)/9 = 343/9. Step 4: Substitute a^2 and b^2 into the standard equation. x^2/7^2 - y^2/(343/9) = 1 x^2/49 - 9y^2/343 = 1 Final answer: The equation of the hyperbola is x^2/49 - 9y^2/343 = 1.
  • Q: Find the equation of the ellipse whose foci are (0, ±3) and a = 5. A: Step 1: Identify the type of ellipse. Since the foci are (0, ±3), they lie on the y-axis, indicating a vertical ellipse. Its standard equation is x^2/b^2 + y^2/a^2 = 1. Step 2: Determine 'c' and 'a'. From foci (0, ±c), we have c = 3. Given a = 5. Step 3: Calculate 'b' using the relationship c^2 = a^2 - b^2. 3^2 = 5^2 - b^2 9 = 25 - b^2 b^2 = 25 - 9 b^2 = 16. Step 4: Substitute a^2 and b^2 into the standard equation. x^2/16 + y^2/5^2 = 1 x^2/16 + y^2/25 = 1 Final answer: The equation of the ellipse is x^2/16 + y^2/25 = 1.

Frequently Asked Questions

What is the main difference between an ellipse and a hyperbola?

The main difference lies in their geometric definition. For an ellipse, the sum of the distances from any point on the curve to two fixed foci is constant. For a hyperbola, the absolute difference of the distances from any point on the curve to two fixed foci is constant. Their equations also differ by a sign: ellipses have a plus sign between the squared terms, while hyperbolas have a minus sign.

How do I determine if an ellipse or hyperbola is horizontal or vertical?

For an ellipse, check where the larger denominator (`a^2`) is. If `a^2` is under `x^2`, it's horizontal. If `a^2` is under `y^2`, it's vertical. For a hyperbola, check which term is positive. If the `x^2` term is positive, it's horizontal. If the `y^2` term is positive, it's vertical. Remember, for a hyperbola, `a^2` is always under the positive term.

What is eccentricity and what does it tell us about conic sections?

Eccentricity (e) is a ratio `c/a` that describes the 'roundness' or 'openness' of a conic section. For an ellipse, `0 < e < 1`; a value closer to 0 means it's more circular, while closer to 1 means it's more elongated. For a hyperbola, `e > 1`; a larger 'e' means the hyperbola branches are wider. A parabola has `e = 1`, and a circle has `e = 0`.

What is the significance of the latus rectum for an ellipse or hyperbola?

The latus rectum is a line segment passing through a focus, perpendicular to the major/transverse axis, and with endpoints on the conic section. Its length (`2b^2/a`) provides another characteristic dimension of the curve. While not always directly used in finding the equation, it is an important property that helps in understanding and sketching the precise shape of the ellipse or hyperbola.