Conic Sections – Exercise 11.3 – Class XI

1. Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the rectum of the ellipse [(x²)/₃₆] + [(y²)/₁₆]= 1

Solution:

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, while the minor axis is along the y-axis

On comparing the given equation with [(x²)/(a²)] + [(y²)/ (b²)]= 1

Therefore, c = √(a² – b²) = √(36-16) = 2√5

We get, a = 6 and b = 4

Thus, the coordinates of the foci are (2√5, 0) and (-2√5, 0)

The coordinates of the vertices are (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 x 16}/₆ = ¹⁶/₃

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2. Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the rectum of the ellipse [(x²)/₄] + [(y²)/₂₅]= 1

Solution:

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, while the minor axis is along the x-axis

On comparing the given equation with [(x²)/(b²)] + [(y²)/ (a²)]= 1

We get, a = 5 and b = 2

c = √(a² – b²) = √(25-4) = √21

Thus, the coordinates of the foci are (0, √21) and (0, -√21)

The coordinates of the vertices are (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 x 4}/₅ = ⁸/₅

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3. Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the rectum of the ellipse [(x²)/₁₆] + [(y²)/₉]= 1

Solution:

The given equation is [(x²)/₁₆] + [(y²)/₉]= 1

Here the denominator of [(x²)/₁₆] is greater than the denominator of [(y²)/₉]

Therefore, the major axis is along the x- axis, while the minor axis is along the y-axis

On comparing the given equation with [(x²)/(a²)] + [(y²)/ (b²)]= 1

We get, a = 4 and b = 3

Therefore, c = √(a² – b²) = √(16-9) =√7

Thus, the coordinates of the foci are (√7, 0) and (-√7, 0)

The coordinates of the vertices are (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 x 9}/₄ = ⁹/₄

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4. Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the rectum of the ellipse [(x²)/₂₅] + [(y²)/₁₀₀]= 1

Solution:

The given equation is [(x²)/₂₅] + [(y²)/₁₀₀]= 1

Here the denominator of [(y²)/₁₀₀] is greater than the denominator of [(x²)/₂₅]

Therefore, the major axis is along the y- axis, while the minor axis is along the x-axis

On comparing the given equation with [(x²)/(b²)] + [(y²)/ (a²)]= 1

Therefore, c = √(a² – b²) = √(100-25) = √75 = 5√3

We get, a = 5 and b = 10

Thus, the coordinates of the foci are (0, 5√3) and (0, -5√3)

The coordinates of the vertices are (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 x 25}/₁₀ =5

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5. Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the rectum of the ellipse [(x²)/₄₉] + [(y²)/₃₆]= 1

Solution:

The given equation is [(x²)/₄₉] + [(y²)/₃₆]= 1

Here the denominator of [(x²)/₄₉] is greater than the denominator of [(y²)/₃₆]

Therefore, the major axis is along the x- axis, while the minor axis is along the y-axis

On comparing the given equation with [(x²)/(a²)] + [(y²)/ (b²)]= 1

We get, a = 7 and b = 6

Therefore, c = √(a² – b²) = √(49-36) =√13

Thus, the coordinates of the foci are (√13, 0) and (-√13, 0)

The coordinates of the vertices are (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 x 36}/₇ = ⁷²/₇

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6. Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the rectum of the ellipse [(x²)/₁₀₀] + [(y²)/₄₀₀]= 1

Solution:

The given equation is [(x²)/₁₀₀] + [(y²)/₄₀₀]= 1

Here the denominator of [(y²)/₄₀₀] is greater than the denominator of [(x²)/₁₀₀]

Therefore, the major axis is along the y- axis, while the minor axis is along the x-axis

On comparing the given equation with [(x²)/(b²)] + [(y²)/ (a²)]= 1

We get, a = 10 and b = 20

Therefore, c = √(a² – b²) = √(100 – 400) =√300 = ±10√3

Thus, the coordinates of the foci are (±10√3, 0) and (-10√3, 0)

The coordinates of the vertices are (20, 0) and (-20, 0)

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 x 100}/₂₀ = 10

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7. Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the rectum of the ellipse 36x² + 4y²= 1

Solution:

The given equation is 36x² + 4y²= 1

36x² + 4y²= 1 can also be written as [(x²)/₄] + [(y²)/₃₆] = 1

i.e., [(x²)/(2²)] + [(y²)/(6²)] = 1

Here the denominator of [(y²)/6²] is greater than the denominator of [(x²)/2²]

Therefore, the major axis is along the y- axis, while the minor axis is along the x-axis

On comparing the given equation with [(x²)/(b²)] + [(y²)/ (a²)]= 1

We get, a = 2 and b = 6

Therefore, c = √(a² – b²) = √(36-4) =√32 = 4√2

Thus, the coordinates of the foci are (4√2, 0) and (-4√2, 0)

The coordinates of the vertices are (6, 0) and (-6, 0)

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 x 4}/₆ = ⁴/₃

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8. Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the rectum of the ellipse 16x² + y²= 16

Solution:

The given equation is 16x² + y²= 16

16x² + y²= 16 can also be written as [(x²)/₁] + [(y²)/₁₆] = 1

i.e., [(x²)/(1²)] + [(y²)/(4²)] = 1

Here the denominator of [(y²)/4²] is greater than the denominator of [(x²)/1²]

Therefore, the major axis is along the y- axis, while the minor axis is along the x-axis

On comparing the given equation with [(x²)/(b²)] + [(y²)/ (c²)]= 1

We get, a = 4 and b = 1

Therefore, c = √(a² – b²) = √(16-1) =√15

Thus, the coordinates of the foci are (√15, 0) and (-√15, 0)

The coordinates of the vertices are (4, 0) and (-4, 0)

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 x 1}/₄ = ¹/₂

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9. Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the rectum of the ellipse 4x² + 9y²= 36

Solution:

The given equation is 4x² + 9y²= 36

4x² + 9y²= 36 can also be written as [(x²)/₉] + [(y²)/₄] = 1

i.e., [(x²)/(3²)] + [(y²)/(2²)] = 1

Here the denominator of [(x²)/2²] is greater than the denominator of [(y²)/3²]

Therefore, the major axis is along the x- axis, while the minor axis is along the y-axis

On comparing the given equation with [(x²)/(a²)] + [(y²)/ (b²)]= 1

We get, a = 3 and b = 2

Therefore, c = √(a² – b²) = √(9-4) =√5

Thus, the coordinates of the foci are (√5, 0) and (-√5, 0)

The coordinates of the vertices are (3, 0) and (-3, 0)

Length of major axis = 2a = 6

Length of minor axis = 2b = 4

Eccentricity e = ^c/_a = ^√5/₃ = ^√5/₃

Length of latus rectum = (2b²)/_a = ^{2 x 4}/₃ = ⁸/₃

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10. Find the equation for the ellipse that satisfies the given conditions: Vertices (±5, 0), foci (±4, 0)

Solution:

Vertices (±5, 0), foci (±4, 0)

Here, the vertices are on the x-axis.

Therefore, the equation of the ellipse will be of the form [(x²)/(a²)] + [(y²)/ (b²)]= 1, where a is the semi-major axis.

Accordingly, a = 5 and c = 4.

It is known that a² = b² + c²

5² = b² + 4²

25 = b² + 16

b² = 25 – 16

b = √9 = 3

Thus, the equation of the ellipse is [(x² )/(5²)]+ [(y² )/(3²)] = 1 or [(x² )/(25)]+ [(y² )/(9)] = 1

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11. Find the equation for the ellipse that satisfies the given conditions: Vertices (0, ±13), foci (0, ±5)

Solution:

given vertices (0, ±13), foci (0, ±5)

Here, the vertices are on the y-axis.

Therefore, the equation of the ellipse will be of the form [(x²)/(a²)] + [(y²)/ (b²)]= 1, where a is the semi-major axis.

Accordingly, a = 13 and c = 5.

It is known that a² = b² + c²

13² = b² + 5²

169 = b² + 25

b² = 169 – 25

b = √144 = 12

Thus, the equation of the ellipse is [(x² )/(12²)]+ [(y² )/(13²)] = 1 or [(x² )/(144)]+ [(y² )/(169)] = 1

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12. Find the equation for the ellipse that satisfies the given conditions: Vertices (±6, 0), foci (±4, 0)

Solution:

given vertices (±6, 0), foci (±4, 0)

Here, the vertices are on the x-axis.

Therefore, the equation of the ellipse will be of the form [(x²)/(a²)] + [(y²)/ (b²)]= 1, where a is the semi-major axis.

Accordingly, a = 6 and c = 4.

It is known that a² = b² + c²

6² = b² + 4²

36 = b² + 16

b² = 36 – 16

b = √20

Thus, the equation of the ellipse is [(x² )/(6²)]+ [(y² )/( √20)²] = 1 or [(x² )/36]+ [(y² )/20] = 1

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13. Find the equation for the ellipse that satisfies the given conditions: Ends of major axis(±3, 0), ends of minor axis (0, ±2)

solution;

given; ends of major axis (±3, 0), ends of minor axis (0, ±2)

Here, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form [(x²)/(a²)] + [(y²)/ (b²)]= 1 , where a is the semi-major axis.

Accordingly, a = 3 and b = 2.

Thus, the equation of the ellipse is

[(x²)/(3²)] + [(y²)/ (2²)]= 1 i.e., (x²)/₉ + (y²)/₄ = 1

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14. Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (0, ±√5), ends of minor axis (±1, 0)

solution:

Given ennds of major axis (0, ±√5), ends of minor axis (±1, 0)

Here, the major axis is along the y-axis.

Therefore, the equation of the ellipse will be of the form [(x²)/(a²)] + [(y²)/ (b²)]= 1, where a is the semi-major axis.

Accordingly, a =√5 and b = 1.

Thus, the equation of the ellipse is

[(x²)/(1²)] + [(y²)/ (√5)²]= 1

i.e.,

[(x²)/1] + [(y²)/ 5]= 1

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15. Find the equation for the ellipse that satisfies the given conditions: Length of major axis 26, foci (±5, 0)

solution:

Length of major axis = 26; foci = (±5, 0).

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form [(x²)/(a²)] + [(y²)/ (b²)]= 1, where a is the semi-major axis.

Accordingly, 2a = 26 ⇒ a = 13 and c = 5.

a² = b² + c²

13² = b² + 5²

b² = 169 – 25

b = √144 = 12

Thus, the equation of the ellipse is [(x²)/(13²)] + [(y²)/ (12²)]= 1 i.e., [(x²)/₁₆₉] + [(y²)/ ₁₄₄]= 1

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16. Find the equation for the ellipse that satisfies the given conditions: Length of minor axis 16, foci (0, ±6)

solution:

Length of minor axis = 16; foci = (0, ±6).

Since the foci are on the y-axis, the major axis is along the y-axis.

Therefore, the equation of the ellipse will be of the form [(x²)/(a²)] + [(y²)/ (b²)]= 1 , where a is the semi-major axis.

Accordingly, 2b = 16 ⇒ b = 8 and c = 6.

it is known that a² = b² + c²

a² = 8²+6²

a² = 64 + 36

a² = 100

a = 10

Thus, the equation of the ellipse is . [(x²)/(8²)] + [(y²)/ (10)²]= 1 or [(x²)/₆₄] + [(y²)/₁₀₀]= 1

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17. Find the equation for the ellipse that satisfies the given conditions: Foci (±3, 0), a = 4

Solution:

Foci (±3, 0), a = 4

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form [(x²)/(a²)] + [(y²)/ (b²)]= 1, where a is the

semi-major axis.

Accordingly, c = 3 and a = 4.

We know, a² = b² + c²

4² = b² + 3²

16 = b² + 9

b² = 16 – 9 = 7

Thus, the equation of the ellipse is [(x²)/(4²)] + [(y²)/ (√7)²]= 1 or [(x²)/₁₆] + [(y²)/₇]= 1

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18. Find the equation for the ellipse that satisfies the given conditions: b = 3, c = 4, centre at the origin; foci on the x axis.

Solution:

It is given that b = 3, c = 4, centre at the origin; foci on the x axis.

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form [(x²)/(b²)] + [(y²)/(a)²]= 1, where a is the semi-major axis.

Accordingly, b = 3, c = 4.

We know that, a² = b² + c²

a² = 3² + 4² = 9 = 16 = 25

a = 5

Thus, the equation of the ellipse is [(x²)/(5²)] + [(y²)/ (3)²]= 1 or [(x²)/₂₅] + [(y²)/₉]= 1

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19. Find the equation for the ellipse that satisfies the given conditions: Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).

solution:

Since the centre is at (0, 0) and the major axis is on the y-axis, the equation of the ellipse will be of the form

[(x²)/(b²)] + [(y²)/(a)²]= 1 ———–(1)

where a is the semi-major axis

The ellipse passes through points (3, 2) and (1, 6).

Hence,

[(9²)/(b²)] + [(4²)/ (a)²]= 1 ——–(2)

[¹/b²] + [³⁶/a²]= 1 —————-(3)

On solving equations (2) and (3), we obtain b² = 10 and a² = 40.

Thus, the equation of the ellipse is [(x²)/10] + [(y²)/40]= 1 or 4x² + y² = 40

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20. Find the equation for the ellipse that satisfies the given conditions: Major axis on the x -axis and passes through the points (4, 3) and (6, 2).

Solution;

Since the major axis is on the x-axis, the equation of the ellipse will be of the form

(x²)/(a²) + (y²)/(b²) = 1 ———–(1), where a is the semi-major axis.

The ellipse passes through points (4, 3) and (6, 2). Hence,

[¹⁶/(a²)] + [⁹/(b)²]= 1 ———–(2)

³⁶/(a²) + ⁴/(b²) = 1 —————(3)

On solving equations (2) and (3), we obtain a² = 52 and b² = 13.

Thus, the equation of the ellipse is (x²)/₅₂ + (y²)/₁₃ = 1 or x² + 4y² = 52

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