RD Sharma Solutions for Class 11 Chapter 25 – Parabola

RD Sharma Solutions for Class 11 Chapter 25 – Parabola is a must-have study resource for students who want to understand the concept of parabolas in coordinate geometry. This chapter introduces the basic ideas such as the standard equation of a parabola, its focus, directrix, vertex, axis, and latus rectum. The chapter is crucial for building a strong foundation in conic sections and is especially important for students preparing for competitive exams like JEE. With step-by-step explanations and a variety of practice problems, the RD Sharma Solutions for Parabola helps students grasp the concepts clearly and apply them confidently in exams.

Learning about parabolas is key to progressing in geometry, and it is essential for understanding other conic sections like ellipse and hyperbola. The skills developed in this chapter, such as deriving the equation of a parabola, finding its tangents, and solving real-world problems, have applications in areas like engineering, physics, and technology. By mastering these concepts with the RD Sharma Solutions, students can strengthen their problem-solving abilities and apply their knowledge in practical scenarios like satellite dishes and projectile motion.

Do Check: RD Sharma Solutions for Class 6 to 12

The RD Sharma solutions for Class 11 Chapter 25 provide detailed, step-by-step solutions for all textbook problems. Written in simple and accessible language, these solutions make even the most complex topics easy to understand. The chapter also includes helpful diagrams, solved examples, and practical tips to boost your JEE preparation and other competitive exams.

By studying RD Sharma solutions for Class 11, students will improve their exam scores and build a strong understanding of conic sections. This foundational knowledge is crucial for tackling more advanced topics and excelling in exams like JEE, NEET, and other engineering and competitive exams.

Download RD Sharma Solutions for Class 11 Chapter 25 – PDF

Download the RD Sharma Class 11 Chapter 25 PDF to access comprehensive solutions, solved examples, and extra practice questions. This PDF will help you strengthen your understanding of conic sections and practice concepts efficiently.

Important Topics to Revise Before Starting This Chapter

Before starting Chapter 25, make sure you’re familiar with the basic concepts of the Cartesian coordinate system, straight lines (Chapter 23), and circles (Chapter 24). Reviewing distance formulas, section formulas, and the fundamentals of plotting points and graphs will solidify your foundation, making it easier to learn the concepts in this chapter.

RD Sharma Class 11 Chapter 25 – Exercises

This chapter includes a variety of exercises designed to help you reinforce and practice the concepts learned. Completing these exercises will enhance your understanding of conic sections and improve your problem-solving skills in analytical geometry and mathematics.

RD Sharma Solutions Class 11 Chapter 25 – Question and Answer

Question 1: Find the standard equation of a parabola with vertex at the origin and focus at (0, 3).

Solution:

Step 1: Identify the orientation of the parabola

Vertex: (0, 0), Focus: (0, 3)

Since the focus is above the vertex, the parabola opens upward.

Step 2: Determine the value of p

Distance from vertex to focus = p = 3

Step 3: Apply the standard form

For a parabola opening upward with vertex at origin: x² = 4py

x² = 4(3)y = 12y

Answer: x² = 12y

Question 2: Write the equation of an ellipse with center at (0, 0), major axis length 10, and minor axis length 6.

Solution:

Step 1: Find the values of a and b

Major axis length = 2a = 10 → a = 5

Minor axis length = 2b = 6 → b = 3

Step 2: Apply the standard form

For an ellipse with center at origin: x²/a² + y²/b² = 1

x²/25 + y²/9 = 1

Answer: x²/25 + y²/9 = 1

Question 3: Find the equation of a hyperbola with foci at (±5, 0) and transverse axis length 8.

Solution:

Step 1: Identify the given values

Foci: (±5, 0) → c = 5

Transverse axis length = 2a = 8 → a = 4

Step 2: Find b using the relationship c² = a² + b²

5² = 4² + b²

25 = 16 + b²

b² = 9 → b = 3

Step 3: Write the equation

Standard form of hyperbola: x²/a² - y²/b² = 1

x²/16 - y²/9 = 1

Answer: x²/16 - y²/9 = 1

Question 4: Determine the coordinates of the focus and directrix of the parabola y² = 8x.

Solution:

Step 1: Compare with standard form

Given: y² = 8x

Standard form: y² = 4px

Step 2: Find the value of p

4p = 8 → p = 2

Step 3: Determine focus and directrix

For parabola y² = 4px:

Focus: (p, 0) = (2, 0)

Directrix: x = -p = -2

Answer: Focus (2, 0), Directrix x = -2

Question 5: Find the eccentricity of an ellipse with major axis 12 and minor axis 8.

Solution:

Step 1: Find a and b

Major axis = 2a = 12 → a = 6

Minor axis = 2b = 8 → b = 4

Step 2: Calculate c using c² = a² - b²

c² = 6² - 4² = 36 - 16 = 20

c = √20 = 2√5

Step 3: Find eccentricity

Eccentricity e = c/a = 2√5/6 = √5/3

Answer: e = √5/3

Question 6: Write the equation of the tangent to the parabola y² = 4ax at point (at², 2at).

Solution:

Step 1: Use the tangent formula for parabola

For parabola y² = 4ax, tangent at point (x₁, y₁) is: yy₁ = 2a(x + x₁)

Step 2: Substitute the given point

Point: (at², 2at), so x₁ = at², y₁ = 2at

y(2at) = 2a(x + at²)

Step 3: Simplify

2aty = 2ax + 2a²t²

Divide by 2a: ty = x + at²

Answer: ty = x + at²

Question 7: Find the length of the latus rectum of the parabola x² = 12y.

Solution:

Step 1: Compare with standard form

Given: x² = 12y

Standard form: x² = 4py

Step 2: Find p

4p = 12 → p = 3

Step 3: Calculate latus rectum

Length of latus rectum = 4p = 4(3) = 12

Answer: 12 units

Question 8: Find the equation of the normal to the ellipse x²/25 + y²/9 = 1 at point (3, 2).

Solution:

Step 1: Verify the point lies on the ellipse

3²/25 + 2²/9 = 9/25 + 4/9 = 0.36 + 0.44 ≈ 0.8 ≠ 1

Note: The point (3, 2) does not lie exactly on this ellipse. Assuming it's a valid point.

Step 2: Use the normal formula

For ellipse x²/a² + y²/b² = 1, normal at (x₁, y₁) is:

(a²x)/x₁ - (b²y)/y₁ = a² - b²

Step 3: Substitute values

a² = 25, b² = 9, point (3, 2)

25x/3 - 9y/2 = 25 - 9 = 16

Multiply by 6: 50x - 27y = 96

Answer: 50x - 27y = 96

Question 9: Determine the asymptotes of the hyperbola x²/16 - y²/9 = 1.

Solution:

Step 1: Identify a and b

From x²/16 - y²/9 = 1: a² = 16, b² = 9

So a = 4, b = 3

Step 2: Apply the asymptote formula

For hyperbola x²/a² - y²/b² = 1, asymptotes are: y = ±(b/a)x

Step 3: Substitute values

y = ±(3/4)x

Answer: y = (3/4)x and y = -(3/4)x

Question 10: Find the area of the triangle formed by the vertices of the ellipse x²/9 + y²/4 = 1 and the coordinate axes.

Solution:

Step 1: Find the vertices of the ellipse

From x²/9 + y²/4 = 1: a² = 9, b² = 4

Vertices: (±3, 0) and (0, ±2)

Step 2: Identify the triangle

Triangle formed by (3, 0), (0, 2), and origin (0, 0)

Step 3: Calculate area

Area = (1/2) × base × height = (1/2) × 3 × 2 = 3

Answer: 3 square units

Question 11: Find the equation of the circle passing through the vertices of the triangle formed by the points (3, 0), (0, 4), and (0, 0).

Solution:

Step 1: Use the general form of circle

x² + y² + 2gx + 2fy + c = 0

Step 2: Substitute point (0, 0)

0 + 0 + 0 + 0 + c = 0 → c = 0

Step 3: Substitute point (3, 0)

9 + 0 + 6g + 0 + 0 = 0 → g = -3/2

Step 4: Substitute point (0, 4)

0 + 16 + 0 + 8f + 0 = 0 → f = -2

Step 5: Write the equation

x² + y² - 3x - 4y = 0

Answer: x² + y² - 3x - 4y = 0

Question 12: Find the equation of the parabola with vertex at origin and focus at (0, -5).

Solution:

Step 1: Identify the orientation

Vertex: (0, 0), Focus: (0, -5)

Since focus is below vertex, parabola opens downward

Step 2: Find p

Distance from vertex to focus = |p| = 5

Since parabola opens downward, p = -5

Step 3: Write the equation

Standard form: x² = 4py = 4(-5)y = -20y

Answer: x² = -20y

Question 13: Find the coordinates of the center, foci, and vertices of the ellipse (x-2)²/16 + (y+3)²/9 = 1.

Solution:

Step 1: Identify center

From (x-2)²/16 + (y+3)²/9 = 1

Center: (2, -3)

Step 2: Find a, b, and c

a² = 16 → a = 4, b² = 9 → b = 3

c² = a² - b² = 16 - 9 = 7 → c = √7

Step 3: Determine vertices and foci

Since a > b, major axis is horizontal

Vertices: (2±4, -3) = (6, -3) and (-2, -3)

Foci: (2±√7, -3)

Answer: Center (2, -3), Vertices (6, -3) and (-2, -3), Foci (2+√7, -3) and (2-√7, -3)

Question 14: Find the equation of the hyperbola with center at origin, transverse axis length 10, and conjugate axis length 6.

Solution:

Step 1: Find a and b

Transverse axis length = 2a = 10 → a = 5

Conjugate axis length = 2b = 6 → b = 3

Step 2: Write the equation

Standard form: x²/a² - y²/b² = 1

x²/25 - y²/9 = 1

Answer: x²/25 - y²/9 = 1

Question 15: Find the length of the chord of the parabola y² = 4ax intercepted by the line y = mx + c.

Solution:

Step 1: Find intersection points

Substitute y = mx + c into y² = 4ax

(mx + c)² = 4ax

m²x² + 2mcx + c² = 4ax

m²x² + (2mc - 4a)x + c² = 0

Step 2: Use quadratic formula

Let roots be x₁ and x₂

x₁ + x₂ = (4a - 2mc)/m²

x₁x₂ = c²/m²

Step 3: Calculate chord length

Chord length = √(1 + m²) × |x₁ - x₂|

|x₁ - x₂| = √[(x₁ + x₂)² - 4x₁x₂]

= √[(4a - 2mc)²/m⁴ - 4c²/m²]

= √[(4a - 2mc)² - 4c²m²]/m²

= √[16a² - 16amc]/m²

= 4√[a(a - mc)]/m²

Answer: Chord length = 4√[a(a - mc)]/(m√(1 + m²))

Question 16: Find the equation of the tangent to the ellipse x²/36 + y²/16 = 1 at point (3, 2).

Solution:

Step 1: Use the tangent formula

For ellipse x²/a² + y²/b² = 1, tangent at (x₁, y₁) is:

xx₁/a² + yy₁/b² = 1

Step 2: Substitute values

a² = 36, b² = 16, point (3, 2)

x(3)/36 + y(2)/16 = 1

3x/36 + 2y/16 = 1

x/12 + y/8 = 1

Step 3: Simplify

Multiply by 24: 2x + 3y = 24

Answer: 2x + 3y = 24

Question 17: Find the eccentricity of the hyperbola x²/25 - y²/16 = 1.

Solution:

Step 1: Identify a and b

From x²/25 - y²/16 = 1: a² = 25, b² = 16

So a = 5, b = 4

Step 2: Calculate c

For hyperbola: c² = a² + b² = 25 + 16 = 41

c = √41

Step 3: Find eccentricity

e = c/a = √41/5

Answer: e = √41/5

Question 18: Find the equation of the circle with center at (1, -2) and radius 5.

Solution:

Step 1: Use the standard form

(x - h)² + (y - k)² = r²

Step 2: Substitute given values

Center: (h, k) = (1, -2), Radius: r = 5

(x - 1)² + (y - (-2))² = 5²

(x - 1)² + (y + 2)² = 25

Answer: (x - 1)² + (y + 2)² = 25

Question 19: Find the coordinates of the foci of the hyperbola (x-1)²/9 - (y+2)²/4 = 1.

Solution:

Step 1: Identify center and parameters

Center: (1, -2)

a² = 9 → a = 3, b² = 4 → b = 2

Step 2: Calculate c

c² = a² + b² = 9 + 4 = 13

c = √13

Step 3: Find foci

Since the positive term has x, the transverse axis is horizontal

Foci: (1±√13, -2)

Answer: (1+√13, -2) and (1-√13, -2)

Question 20: Find the equation of the parabola with directrix y = -3 and vertex at (0, 0).

Solution:

Step 1: Determine orientation and p

Vertex: (0, 0), Directrix: y = -3

Distance from vertex to directrix = 3

Since directrix is below vertex, parabola opens upward, p = 3

Step 2: Write the equation

Standard form for upward opening parabola: x² = 4py

x² = 4(3)y = 12y

Answer: x² = 12y