NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 Conic Sections 2026-27

In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of the major axis, the length of the minor axis, the eccentricity, and the length of the latus rectum of the

Ellipse.

Question 1: x²/36 + y²/16 = 1

Ans: The given equation is x²/36 + y²/16 = 1.

Here, the denominator of x²/36 is greater than the denominator of y²/16. Therefore, the major axis is along the x-axis, and the minor axis is along the y-axis.

Comparing with the standard equation x²/a² + y²/b² = 1, we get a² = 36 and b² = 16, so a = 6 and b = 4.

Now, c = √(a² − b²) = √(36 − 16) = √20 = 2√5

• Coordinates of the foci: (2√5, 0) and (−2√5, 0)

• Coordinates of the vertices: (6, 0) and (−6, 0)

• Length of major axis = 2a = 12

• Length of minor axis = 2b = 8

• Eccentricity, e = c/a = 2√5/6 = √5/3

• Length of latus rectum = 2b²/a = (2 × 16)/6 = 16/3

Question 2: x²/4 + y²/25 = 1

Ans: The given equation is x²/4 + y²/25 = 1.

Here, the denominator of y²/25 is greater than the denominator of x²/4. Therefore, the major axis is along the y-axis, and the minor axis is along the x-axis.

Comparing with x²/b² + y²/a² = 1, we get a² = 25 and b² = 4, so a = 5 and b = 2.

Now, c = √(a² − b²) = √(25 − 4) = √21

• Coordinates of the foci: (0, √21) and (0, −√21)

• Coordinates of the vertices: (0, 5) and (0, −5)

• Length of major axis = 2a = 10

• Length of minor axis = 2b = 4

• Eccentricity, e = c/a = √21/5

• Length of latus rectum = 2b²/a = (2 × 4)/5 = 8/5

Question 3: x²/16 + y²/9 = 1

Ans: The denominator of x²/16 is greater than that of y²/9, so the major axis is along the x-axis.

Comparing with x²/a² + y²/b² = 1, we get a = 4 and b = 3.

Now, c = √(a² − b²) = √(16 − 9) = √7

• Coordinates of the foci: (√7, 0) and (−√7, 0)

• Coordinates of the vertices: (4, 0) and (−4, 0)

• Length of major axis = 2a = 8

• Length of minor axis = 2b = 6

• Eccentricity, e = c/a = √7/4

• Length of latus rectum = 2b²/a = (2 × 9)/4 = 9/2

Question 4: x²/25 + y²/100 = 1

Ans: The denominator of y²/100 is greater than that of x²/25, so the major axis is along the y-axis.

Comparing with x²/b² + y²/a² = 1, we get a = 10 and b = 5.

Now, c = √(a² − b²) = √(100 − 25) = √75 = 5√3

• Coordinates of the foci: (0, 5√3) and (0, −5√3)

• Coordinates of the vertices: (0, 10) and (0, −10)

• Length of major axis = 2a = 20

• Length of minor axis = 2b = 10

• Eccentricity, e = c/a = 5√3/10 = √3/2

• Length of latus rectum = 2b²/a = (2 × 25)/10 = 5

Question 5: x²/49 + y²/36 = 1

Ans: The denominator of x²/49 is greater than that of y²/36, so the major axis is along the x-axis.

Comparing with x²/a² + y²/b² = 1, we get a = 7 and b = 6.

Now, c = √(a² − b²) = √(49 − 36) = √13

• Coordinates of the foci: (√13, 0) and (−√13, 0)

• Coordinates of the vertices: (7, 0) and (−7, 0)

• Length of major axis = 2a = 14

• Length of minor axis = 2b = 12

• Eccentricity, e = c/a = √13/7

• Length of latus rectum = 2b²/a = (2 × 36)/7 = 72/7

Question 6: x²/100 + y²/400 = 1

Ans: The denominator of y²/400 is greater than that of x²/100, so the major axis is along the y-axis.

Comparing with x²/b² + y²/a² = 1, we get a = 20 and b = 10.

Now, c = √(a² − b²) = √(400 − 100) = √300 = 10√3

• Coordinates of the foci: (0, 10√3) and (0, −10√3)

• Coordinates of the vertices: (0, 20) and (0, −20)

• Length of major axis = 2a = 40

• Length of minor axis = 2b = 20

• Eccentricity, e = c/a = 10√3/20 = √3/2

• Length of latus rectum = 2b²/a = (2 × 100)/20 = 10

Question 7: 36x² + 4y² = 144

Ans: Dividing the entire equation by 144, we get:

36x²/144 + 4y²/144 = 1, which gives x²/4 + y²/36 = 1

The denominator of y²/36 is greater than that of x²/4, so the major axis is along the y-axis.

Comparing with x²/b² + y²/a² = 1, we get a = 6 and b = 2.

Now, c = √(a² − b²) = √(36 − 4) = √32 = 4√2

• Coordinates of the foci: (0, 4√2) and (0, −4√2)

• Coordinates of the vertices: (0, 6) and (0, −6)

• Length of major axis = 2a = 12

• Length of minor axis = 2b = 4

• Eccentricity, e = c/a = 4√2/6 = 2√2/3

• Length of latus rectum = 2b²/a = (2 × 4)/6 = 4/3

Question 8: 16x² + y² = 16

Ans: Dividing the entire equation by 16, we get:

x²/1 + y²/16 = 1

The denominator of y²/16 is greater than that of x²/1, so the major axis is along the y-axis.

Comparing with x²/b² + y²/a² = 1, we get a = 4 and b = 1.

Now, c = √(a² − b²) = √(16 − 1) = √15

• Coordinates of the foci: (0, √15) and (0, −√15)

• Coordinates of the vertices: (0, 4) and (0, −4)

• Length of major axis = 2a = 8

• Length of minor axis = 2b = 2

• Eccentricity, e = c/a = √15/4

• Length of latus rectum = 2b²/a = (2 × 1)/4 = ½

Question 9: 4x² + 9y² = 36

Ans: Dividing the entire equation by 36, we get:

x²/9 + y²/4 = 1

The denominator of x²/9 is greater than that of y²/4, so the major axis is along the x-axis.

Comparing with x²/a² + y²/b² = 1, we get a = 3 and b = 2.

Now, c = √(a² − b²) = √(9 − 4) = √5

• Coordinates of the foci: (√5, 0) and (−√5, 0)

• Coordinates of the vertices: (3, 0) and (−3, 0)

• Length of major axis = 2a = 6

• Length of minor axis = 2b = 4

• Eccentricity, e = c/a = √5/3

• Length of latus rectum = 2b²/a = (2 × 4)/3 = 8/3

In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions:

Question 10: Vertices (±5, 0), foci (±4, 0)

Ans: Since the vertices are on the x-axis, the major axis is along the x-axis, and the equation of the ellipse is of the form x²/a² + y²/b² = 1.

From the vertices (±5, 0), we get a = 5. From the foci (±4, 0), we get c = 4.

Using the relation c² = a² − b²: 16 = 25 − b² b² = 25 − 16 = 9

Therefore, the equation of the ellipse is x²/25 + y²/9 = 1

Question 11: Vertices (0, ±13), foci (0, ±5)

Ans: Since the vertices are on the y-axis, the major axis is along the y-axis, and the equation is of the form x²/b² + y²/a² = 1.

From the vertices (0, ±13), we get a = 13. From the foci (0, ±5), we get c = 5.

Using c² = a² − b²: 25 = 169 − b² b² = 169 − 25 = 144

Therefore, the equation of the ellipse is x²/144 + y²/169 = 1

Question 12: Vertices (±6, 0), foci (±4, 0)

Ans: Since the vertices are on the x-axis, the equation is of the form x²/a² + y²/b² = 1.

Here, a = 6 and c = 4.

Using c² = a² − b²: 16 = 36 − b² b² = 36 − 16 = 20

Therefore, the equation of the ellipse is x²/36 + y²/20 = 1

Question 13: Ends of major axis (±3, 0), ends of minor axis (0, ±2)

Ans: Since the ends of the major axis lie on the x-axis, the major axis is along the x-axis, and the equation is of the form x²/a² + y²/b² = 1.

From the ends of the major axis (±3, 0), we get a = 3. From the ends of the minor axis (0, ±2), we get b = 2.

Therefore, the equation of the ellipse is x²/9 + y²/4 = 1

Question 14: Ends of major axis (0, ±√5), ends of minor axis (±1, 0)

Ans: Since the ends of the major axis lie on the y-axis, the major axis is along the y-axis, and the equation is of the form x²/b² + y²/a² = 1.

From the ends of the major axis (0, ±√5), we get a = √5, so a² = 5. From the ends of the minor axis (±1, 0), we get b = 1, so b² = 1.

Therefore, the equation of the ellipse is x²/1 + y²/5 = 1

Question 15: Length of major axis 26, foci (±5, 0)

Ans: Since the foci lie on the x-axis, the major axis is along the x-axis, and the equation is of the form x²/a² + y²/b² = 1.

Given, length of major axis = 2a = 26, so a = 13. From the foci (±5, 0), we get c = 5.

Using c² = a² − b²: 25 = 169 − b² b² = 144

Therefore, the equation of the ellipse is x²/169 + y²/144 = 1

Question 16: Length of minor axis 16, foci (0, ±6)

Ans: Since the foci lie on the y-axis, the major axis is along the y-axis, and the equation is of the form x²/b² + y²/a² = 1.

Given, length of minor axis = 2b = 16, so b = 8. From the foci (0, ±6), we get c = 6.

Using c² = a² − b²: 36 = a² − 64 a² = 36 + 64 = 100

Therefore, the equation of the ellipse is x²/64 + y²/100 = 1

Question 17: Foci (±3, 0), a = 4

Ans: Since the foci lie on the x-axis, the major axis is along the x-axis, and the equation is of the form x²/a² + y²/b² = 1.

Given, c = 3 and a = 4.

Using c² = a² − b²: 9 = 16 − b² b² = 16 − 9 = 7

Therefore, the equation of the ellipse is x²/16 + y²/7 = 1

Question 18: b = 3, c = 4, centre at the origin; foci on the x-axis

Ans: Since the foci lie on the x-axis, the major axis is along the x-axis, and the equation is of the form x²/a² + y²/b² = 1.

Given, b = 3 and c = 4.

Using the relation a² = b² + c²: a² = 9 + 16 = 25

Therefore, the equation of the ellipse is x²/25 + y²/9 = 1

Question 19: Centre at (0, 0), major axis on the y-axis, and passes through the points (3, 2) and (1, 6)

Ans: Since the major axis is along the y-axis, the equation of the ellipse is of the form:

x²/b² + y²/a² = 1 ... (i)

Since the ellipse passes through the point (3, 2), substituting in equation (i): 9/b² + 4/a² = 1 ... (1)

Since the ellipse passes through the point (1, 6), substituting in equation (i): 1/b² + 36/a² = 1 ... (2)

Let 1/b² = p and 1/a² = q. Then the equations become: 9p + 4q = 1 ... (3) p + 36q = 1 ... (4)

From equation (4): p = 1 − 36q

Substituting in equation (3): 9(1 − 36q) + 4q = 1 9 − 324q + 4q = 1 −320q = −8 q = 1/40, which gives a² = 40

Now, p = 1 − 36 × (1/40) = 1 − 36/40 = 4/40 = 1/10, which gives b² = 10

Therefore, the equation of the ellipse is x²/10 + y²/40 = 1

Question 20: The major axis on the x-axis and passes through the points (4, 3) and (6, 2)

Ans: Since the major axis is along the x-axis, the equation of the ellipse is of the form:

x²/a² + y²/b² = 1 ... (i)

Since the ellipse passes through the point (4, 3): 16/a² + 9/b² = 1 ... (1)

Since the ellipse passes through the point (6, 2): 36/a² + 4/b² = 1 ... (2)

Let 1/a² = p and 1/b² = q. Then: 16p + 9q = 1 ... (3) 36p + 4q = 1 ... (4)

Multiplying equation (3) by 4 and equation (4) by 9: 64p + 36q = 4 ... (5) 324p + 36q = 9 ... (6)

Subtracting equation (5) from equation (6): 260p = 5 p = 1/52, which gives a² = 52

Substituting p = 1/52 in equation (3): 16/52 + 9q = 1 9q = 1 − 4/13 = 9/13 q = 1/13, which gives b² = 13

Therefore, the equation of the ellipse is x²/52 + y²/13 = 1

Key Takeaways from NCERT Solutions Class 11 Maths Chapter 10 Exercise 10.3 Conic Sections

Understanding the concepts covered in NCERT Solutions Class 11 Maths Chapter 10 Exercise 10.3 Conic Sections helps students develop a strong foundation in coordinate geometry. This exercise focuses on the properties, equations, and applications of conic sections, making it an important topic for both school exams and competitive examinations.

Make sure to cover key concepts such as the standard equations of circles, parabolas, ellipses, and hyperbolas, along with their geometric interpretations. Regularly practising NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 improves problem-solving skills and boosts confidence for the 2026–27 academic session.

Practice solving questions step by step and focus on understanding the underlying concepts rather than memorising formulas. This approach enhances mathematical reasoning, improves accuracy, and helps students perform well in examinations.

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