Conic Sections

Chapter 10.1: Introduction

• Conic Section: The curve formed by the intersection of a plane with a double cone.
• Types of Conic Sections: Circle, ellipse, parabola, and hyperbola.

Chapter 10.2: The Circle

• Equation of a Circle: (x – h)^2 + (y – k)^2 = r^2, where (h, k) is the center and r is the radius.
• Standard Form: x^2 + y^2 + 2gx + 2fy + c = 0
• Center and Radius: Center is (-g, -f) and radius is √(g^2 + f^2 – c).

Chapter 10.3: The Parabola

• Equation of a Parabola: y^2 = 4ax (vertical axis) or x^2 = 4ay (horizontal axis)
• Vertex: (0, 0)
• Axis: Line y = 0 (vertical axis) or x = 0 (horizontal axis)
• Focus: (a, 0) or (0, a)
• Directrix: x = -a or y = -a

Chapter 10.4: The Ellipse

• Equation of an Ellipse: x^2/a^2 + y^2/b^2 = 1 (standard form)
• Center: (0, 0)
• Vertices: (±a, 0) and (0, ±b)
• Foci: (±√(a^2 – b^2), 0) or (0, ±√(a^2 – b^2))
• Eccentricity: e = √(a^2 – b^2) / a

Chapter 10.5: The Hyperbola

• Equation of a Hyperbola: x^2/a^2 – y^2/b^2 = 1 (horizontal axis) or y^2/b^2 – x^2/a^2 = 1 (vertical axis)
• Center: (0, 0)
• Vertices: (±a, 0) or (0, ±b)
• Foci: (±√(a^2 + b^2), 0) or (0, ±√(a^2 + b^2))
• Eccentricity: e = √(a^2 + b^2) / a
• Asymptotes: y = ±(b/a)x

Key Concepts:

• Properties of conic sections
• Equations of conic sections
• Graphs of conic sections
• Applications of conic sections (e.g., optics, astronomy)

Exercise 10.1

In each of the following Exercises 1 to 5, find the equation of the circle with

1. centre (0,2) and radius 2

Ans:  

(x – h)^2 + (y – k)^2 = r^2

In this case, the center is (0, 2) and the radius is 2. Substituting these values into the equation:  

(x – 0)^2 + (y – 2)^2 = 2^2

Simplifying:

x^2 + y^2 – 4y + 4 = 4

x^2 + y^2 – 4y = 0

2. centre (–2,3) and radius 4

Ans : 

(x – h)^2 + (y – k)^2 = r^2

Substituting these values into the equation:  

(x – (-2))^2 + (y – 3)^2 = 4^2

(x + 2)^2 + (y – 3)^2 = 16

Expanding the equation:

x^2 + 4x + 4 + y^2 – 6y + 9 = 16

x^2 + y^2 + 4x – 6y – 3 = 0

3. Centre (1/2, 1/4 ) and radius 1/12

Ans : 

(x – h)^2 + (y – k)^2 = r^2

In this case, the center is (1/2, 1/4) and the radius is 1/12. Substituting these values into the equation:  

(x – 1/2)^2 + (y – 1/4)^2 = (1/12)^2

Expanding the equation:

x^2 – x + 1/4 + y^2 – y/2 + 1/16 = 1/144

Multiplying both sides by 144:

144x^2 – 144x + 36 + 144y^2 – 72y + 9 = 1

144x^2 + 144y^2 – 144x – 72y + 44 = 0

4. Centre (1,1) and radius 2

Ans : 

(x – h)^2 + (y – k)^2 = r^2

In this case, the center is (1, 1) and the radius is 2. Substituting these values into the equation:  

(x – 1)^2 + (y – 1)^2 = 2^2

Expanding the equation:

x^2 – 2x + 1 + y^2 – 2y + 1 = 4

x^2 + y^2 – 2x – 2y – 2 = 0

5. centre (–a, –b) and radius a2–b2

Ans : 

In each of the following Exercises 6 to 9, find the centre and radius of the circles.

6. (x + 5)2 + (y – 3)2 = 36

Ans : 

The given equation is already in the standard form of a circle:

(x – h)^2 + (y – k)^2 = r^2

where (h, k) is the center of the circle and r is its radius.

Comparing the given equation with the standard form, we can see that:  

• h = -5
• k = 3
• r^2 = 36

Therefore, the center of the circle is (-5, 3) and its radius is √36 = 6 units.

7. x 2 + y 2 – 4x – 8y – 45 = 0

Ans : 

The given equation is:

x^2 + y^2 – 4x – 8y – 45 = 0

(x – h)^2 + (y – k)^2 = r^2

To do this, we can complete the square for both the x and y terms:  

x^2 – 4x + 4 + y^2 – 8y + 16 – 45 = 0 + 4 + 16

(x – 2)^2 + (y – 4)^2 – 25 = 0

(x – 2)^2 + (y – 4)^2 = 25

Comparing this equation to the standard form, we can see that:

• h = 2
• k = 4
• r^2 = 25

Therefore, the center of the circle is (2, 4) and its radius is √25 = 5 units.

8. x 2 + y 2 – 8x + 10y – 12 = 0

Ans : 

The given equation is:

x^2 + y^2 – 8x + 10y – 12 = 0

(x – h)^2 + (y – k)^2 = r^2

To do this, we can complete the square for both the x and y terms:  

x^2 – 8x + 16 + y^2 + 10y + 25 – 12 = 0 + 16 + 25

(x – 4)^2 + (y + 5)^2 – 43 = 0

(x – 4)^2 + (y + 5)^2 = 43

• h = 4
• k = -5
• r^2 = 43

Therefore, the center of the circle is (4, -5) and its radius is √43 units.

9. 2x 2 + 2y 2 – x = 0

Ans : 

To find the center and radius of the circle represented by the equation 2x^2 + 2y^2 – x = 0, we need to convert it into the standard form of a circle’s equation:

(x – h)^2 + (y – k)^2 = r^2

Steps:

1. Divide the entire equation by 2 to simplify:
   • x^2 + y^2 – x/2 = 0
2. Rearrange to group the x terms:
   • x^2 – x/2 + y^2 = 0
3. Complete the square for the x terms:
   • To complete the square, add and subtract (1/4)^2 (the square of half the coefficient of x):
     • x^2 – x/2 + (1/4)^2 – (1/4)^2 + y^2 = 0
4. Rewrite the equation:
   • (x – 1/4)^2 + y^2 = (1/4)^2

Now, the equation is in standard form.

Center: (1/4, 0) Radius: √(1/4)^2 = 1/4

Therefore, the center of the circle is (1/4, 0) and its radius is 1/4.

10. Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x + y = 16.

Ans :

11. Find the equation of the circle passing through the points (2,3) and (–1,1) and whose centre is on the line x – 3y – 11 = 0.

Ans : 

Since the circle passes through the points (2, 3) and (-1, 1), we have:

(2 – h)^2 + (3 – k)^2 = (h + 1)^2 + (k – 1)^2

Expanding and simplifying:

h^2 – 4h + k^2 – 6k + 13 = h^2 + 2h + k^2 – 2k + 2

-6h – 4k + 11 = 0

3h + 2k – 11/2 = 0

Since the center lies on the line x – 3y – 11 = 0, we also have:

h – 3k – 11 = 0

Now, we have two equations:

3h + 2k – 11/2 = 0 h – 3k – 11 = 0

Solving these equations simultaneously, we get:

h = 7/2 k = -5/2

Therefore, the center of the circle is (7/2, -5/2).

To find the radius, we can use the distance between the center and any of the given points:

Radius = √((2 – 7/2)^2 + (3 – (-5/2))^2)

Radius = √((-1/2)^2 + (11/2)^2)

Radius = √(122/4)

Radius = √30.5

Finally, the equation of the circle is:

(x – 7/2)^2 + (y + 5/2)^2 = 30.5

12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3). 

Ans : 

Let the center of the circle be (a, 0), since it lies on the x-axis.

The radius of the circle is given as 5 units.

Using the distance formula, we can write the equation:

(x – a)^2 + (y – 0)^2 = 5^2

(x – a)^2 + y^2 = 25

Since the circle passes through the point (2, 3), substituting these values in the equation:

(2 – a)^2 + 3^2 = 25

4 – 4a + a^2 + 9 = 25

a^2 – 4a – 12 = 0

Factoring the equation:

(a – 6)(a + 2) = 0

Therefore, the possible values for a are 6 and -2.

So, there are two circles that satisfy the given conditions:

1. Center: (6, 0), Radius: 5 Equation: (x – 6)^2 + y^2 = 25
2. Center: (-2, 0), Radius: 5 Equation: (x + 2)^2 + y^2 = 25

13. Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

Ans : 

(0 – h)^2 + (0 – k)^2 = r^2

h^2 + k^2 = r^2

This means that the points (a, 0) and (0, b) lie on the circle.

Substituting (a, 0) into the equation of the circle:

(a – h)^2 + (0 – k)^2 = r^2

a^2 – 2ah + h^2 + k^2 = r^2

Since h^2 + k^2 = r^2 (from the first equation), we can substitute r^2:

a^2 – 2ah + r^2 = r^2

a^2 – 2ah = 0

a(a – 2h) = 0

Therefore, either a = 0 or a = 2h.

Similarly, substituting (0, b) into the equation of the circle:

(0 – h)^2 + (b – k)^2 = r^2

h^2 + b^2 – 2bk + k^2 = r^2

Since h^2 + k^2 = r^2, we can substitute r^2:

h^2 + b^2 – 2bk + r^2 = r^2

b^2 – 2bk = 0

b(b – 2k) = 0

Therefore, either b = 0 or b = 2k.

Now, we have four possible cases:

1. a = 0 and b = 0: This means the circle is centered at the origin, and its equation is x^2 + y^2 = r^2. Since it passes through (a, 0), the radius is a. So the equation is x^2 + y^2 = a^2.
2. a = 0 and b = 2k: In this case, the center is (0, k) and the radius is k. The equation of the circle is x^2 + (y – k)^2 = k^2.
3. a = 2h and b = 0: In this case, the center is (h, 0) and the radius is h. The equation of the circle is (x – h)^2 + y^2 = h^2.
4. a = 2h and b = 2k: In this case, the center is (h, k) and the radius is √(h^2 + k^2). The equation of the circle is (x – h)^2 + (y – k)^2 = h^2 + k^2.

Since the circle passes through the origin, the radius must be equal to the distance between the origin and the center. Therefore, in all cases, r = √(h^2 + k^2).

Combining the four cases, we can say that the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes is:

x^2 + y^2 – ax – by = 0

14. Find the equation of a circle with centre (2,2) and passes through the point (4,5).

Ans : 

(x – h)^2 + (y – k)^2 = r^2

In this case, the center is (2, 2). To find the radius, we can use the distance formula between the center and the point (4, 5):  

r = √((4 – 2)^2 + (5 – 2)^2) = √(4 + 9) = √13

(x – 2)^2 + (y – 2)^2 = (√13)^2

(x – 2)^2 + (y – 2)^2 = 13

Expanding the equation:

x^2 – 4x + 4 + y^2 – 4y + 4 = 13

x^2 + y^2 – 4x – 4y – 5 = 0

Therefore, the equation of the circle passing through the point (4, 5) with center (2, 2) is:

x^2 + y^2 – 4x – 4y – 5 = 0

15. Does the point (–2.5, 3.5) lie inside, outside or on the circle x 2 + y 2 = 25? 

Ans : 

First, let’s identify the center and radius of the circle.

The equation x^2 + y^2 = 25 is in the standard form of a circle: (x – h)^2 + (y – k)^2 = r^2, where (h, k) is the center and r is the radius.

Comparing the given equation to the standard form, we can see that:  

• h = 0
• k = 0
• r = √25 = 5

So, the circle has its center at (0, 0) and a radius of 5.

Now, let’s determine the distance between the point (-2.5, 3.5) and the center (0, 0).

The distance formula is:

d = √((x2 – x1)^2 + (y2 – y1)^2)

Substituting the coordinates:

d = √((-2.5 – 0)^2 + (3.5 – 0)^2)

d = √(6.25 + 12.25)

d = √18.5

Since 18.5 is less than 25 (the square of the radius), the point (-2.5, 3.5) lies inside the circle.

Exercise 10.2

In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

1. y2= 12x

Ans : 

General Form of a Parabola:

• y^2 = 4ax

• 4a = 12
• a = 3

Therefore, the parabola opens to the right.

Focus:

• The focus of a right-opening parabola is (a, 0).
• So, the focus is (3, 0).

Axis of the Parabola:

• Since the vertex is (0, 0), the axis of symmetry is the y-axis.
• Equation of the axis: x = 0

Directrix:

• The directrix of a right-opening parabola is a vertical line x = -a.
• So, the equation of the directrix is x 
• = -3.

Length of the Latus Rectum:

• So, the length of the latus rectum is 4 * 3 = 12.

Summary:

• Focus: (3, 0)
• Axis of the Parabola: x = 0
• Equation of the Directrix: x = -3
• Length of the Latus Rectum: 12

2. x2= 6y

Ans : 

3. y2= -8x

Ans :

Key points:

• Axis of symmetry: x-axis
• Vertex: (0, 0)
• Opens left (since the coefficient of x is negative)

To find the focus and directrix:

• 4a = -8 (comparing with the standard form y^2 = 4ax)
• a = -2
• Focus: (-a, 0) = (-2, 0)
• Directrix: x = a = 2

Therefore, the parabola y^2 = -8x has:

• Vertex: (0, 0)
• Axis of symmetry: x-axis
• Focus: (-2, 0)
• Directrix: x = 2

4. x2= -16y

Ans : 

5.  y2= 10x

Ans : 

The given equation is y^2 = 10x. This represents a parabola.

Key points:

• Axis of symmetry: y-axis
• Vertex: (0, 0)
• Opens right (since the coefficient of x is positive)

To find the focus and directrix:

• 4a = 10 (comparing with the standard form y^2 = 4ax)
• a = 5/2
• Focus: (a, 0) = (5/2, 0)
• Directrix: x = -a = -5/2

Therefore, the parabola y^2 = 10x has:

• Vertex: (0, 0)
• Axis of symmetry: y-axis
• Focus: (5/2, 0)
• Directrix: x = -5/2

6. x2= -9y

Ans : 

Key points:

• Axis of symmetry: y-axis
• Vertex: (0, 0)
• Opens downward (since the coefficient of y is negative)

To find the focus and directrix:

• 4a = -9 
• a = -9/4
• Focus: (0, -a) = (0, 9/4)
• Directrix: y = a = -9/4

Therefore, the parabola x^2 = -9y has:

• Vertex: (0, 0)
• Axis of symmetry: y-axis
• Focus: (0, 9/4)
• Directrix: y = -9/4

In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:

7. Focus (6,0); directrix x = – 6

Ans :

Finding the Equation of the Parabola

Given:

• Focus: (6, 0)
• Directrix: x = -6

Analysis:

y^2 = 4ax

Finding the value of a:

• Vertex: ((6 + (-6))/2, 0) = (0, 0)
  
• This distance is the value of a.  
• a = 6
  

Substituting the value of a into the general equation:

y^2 = 4 * 6 * x

Therefore, the equation of the parabola is:

y^2 = 24x

8. Focus (0,–3); directrix y = 3

Ans : 

Given:

• Focus: (0, -3)
• Directrix: y = 3

Analysis:

x^2 = -4ay

Finding the value of a:

• Vertex: ((0 + 0)/2, (-3 + 3)/2) = (0, 0)
  
• This distance is the value of a.  
• a = 3
  

Substituting the value of a into the general equation:

x^2 = -4 * 3 * y

Therefore, the equation of the parabola is:

x^2 = -12y

9.Vertex (0,0); focus (3,0)

Ans : 

Given:

• Vertex: (0, 0)
• Focus: (3, 0)

Analysis:

y^2 = 4ax

Finding the value of a:

• a = 3

Substituting the value of a into the general equation:

y^2 = 4 * 3 * x

Therefore, the equation of the parabola is:

y^2 = 12x

10.Vertex (0,0); focus (–2,0)

Ans : 

Given:

• Vertex: (0, 0)
• Focus: (-2, 0)

Analysis:

y^2 = -4ax

Finding the value of a:

• a = 2

Substituting the value of a into the general equation:

y^2 = -4 * 2 * x

Therefore, the equation of the parabola is:

y^2 = -8x

11.Vertex (0,0) passing through (2,3) and axis is along x-axis

Ans : 

Given:

• Vertex: (0, 0)
• Axis: Along the x-axis
• Passes through the point (2, 3)

Analysis:

• Since the axis is along the x-axis, the parabola is of the form y^2 = 4ax.
• We need to find the value of a.

Substituting the given point (2, 3) into the equation:

3^2 = 4a * 2

9 = 8a

Solving for a:

a = 9/8

Therefore, the equation of the parabola is:

y^2 = 4 * (9/8) * x

Simplifying:

y^2 = 9/2 * x

12. Vertex (0,0), passing through (5,2) and symmetric with respect to y-axis

Ans : 

Given:

• Vertex: (0, 0)
• Passes through the point (5, 2)
• Symmetric with respect to the y-axis

Analysis:

• Since the parabola is symmetric with respect to the y-axis and has its vertex at (0, 0), the general equation of the parabola is of the form x^2 = 4ay.

Substituting the given point (5, 2) into the equation:

5^2 = 4a * 2

25 = 8a

Solving for a:

a = 25 / 8

Substituting the value of a back into the general equation:

x^2 = 4 * (25/8) * y

Simplifying:

x^2 = 25/2 * y

Therefore, the equation of the parabola is:

x^2 = 25/2 * y

Exercise 10.3

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

1.

Ans : 

Analyzing the Ellipse: x^2/36 + y^2/16 = 1

Standard Form of an Ellipse:

(x^2 / a^2) + (y^2 / b^2) = 1

where:

• a^2 = 36
• b^2 = 16

Finding the Vertices:

• Since a^2 > b^2, the major axis is along the x-axis.
• The vertices are (±a, 0).
• a = √36 = 6
• Vertices: (±6, 0)

Finding the Foci:

• c = √(a^2 – b^2) = √(36 – 16) = √20 = 2√5
• The foci are (±c, 0).
• Foci: (±2√5, 0)

Finding the Length of the Major Axis:

• Length of major axis = 2a = 2 * 6 = 12

Finding the Length of the Minor Axis:

• Length of minor axis = 2b
•  = 2 * 4 
• = 8

Finding the Eccentricity:

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

Finding the Length of the Latus Rectum:

• Length of latus rectum = 2b^2 / a = 2 * 16 / 6 

= 16/3

2.

2. 

Ans : 

Analyzing the Ellipse: x^2/4 + y^2/25 = 1

Standard Form of an Ellipse:

(x^2 / a^2) + (y^2 / b^2) = 1

where:

• a^2 = 4
• b^2 = 25

Finding the Vertices:

• Since a^2 < b^2, the major axis is along the y-axis.
• The vertices are (0, ±b).
• b = √25 = 5
• Vertices: (0, ±5)

Finding the Foci:

• c = √(b^2 – a^2) = √(25 – 4) = √21
• The foci are (0, ±c).
• Foci: (0, ±√21)

Finding the Length of the Major Axis:

• Length of major axis = 2b = 2 * 5 = 10

Finding the Length of the Minor Axis:

• Length of minor axis = 2a = 2 * 2 = 4

Finding the Eccentricity:

• Eccentricity (e) = c / b = (√21) / 5

Finding the Length of the Latus Rectum:

Length of latus rectum = 2a^2 / b = 2 * 4 / 5 = 8/5

3.

Ans : 

Analyzing the Ellipse: x^2/16 + y^2/9 = 1

Standard Form of an Ellipse:

The given equation is already in standard form:

(x^2 / a^2) + (y^2 / b^2) = 1

where:

• a^2 = 16
• b^2 = 9

Finding the Vertices:

• Since a^2 > b^2, the major axis is along the x-axis.
• The vertices are (±a, 0).
• a = √16 = 4
• Vertices: (±4, 0)

Finding the Foci:

• c = √(a^2 – b^2) = √(16 – 9) = √7
• The foci are (±c, 0).
• Foci: (±√7, 0)

Finding the Length of the Major Axis:

• Length of major axis = 2a = 2 * 4 = 8

Finding the Length of the Minor Axis:

• Length of minor axis = 2b = 2 * 3 = 6

Finding the Eccentricity:

• Eccentricity (e) = c / a = (√7) / 4

Finding the Length of the Latus Rectum:

Length of latus rectum = 2b^2 / a = 2 * 9 / 4 = 9/2

4.

Ans : 

Analyzing the Ellipse: x^2/25 + y^2/100 = 1

Standard Form of an Ellipse:

(x^2 / a^2) + (y^2 / b^2) = 1

where:

• a^2 = 25
• b^2 = 100

Finding the Vertices:

• Since a^2 < b^2, the major axis is along the y-axis.
• The vertices are (0, ±b).
• b = √100 = 10
• Vertices: (0, ±10)

Finding the Foci:

• c = √(b^2 – a^2) = √(100 – 25) = √75 = 5√3
• The foci are (0, ±c).
• Foci: (0, ±5√3)

Finding the Length of the Major Axis:

• Length of major axis = 2b = 2 * 10 = 20

Finding the Length of the Minor Axis:

• Length of minor axis = 2a = 2 * 5 = 10

Finding the Eccentricity:

• Eccentricity (e) = c / b = (5√3) / 10 = √3 / 2

Finding the Length of the Latus Rectum:

Length of latus rectum = 2a^2 / b = 2 * 25 / 10 = 5

5

Ans : 

Analyzing the Ellipse: x^2/49 + y^2/36 = 1

Standard Form of an Ellipse:

(x^2 / a^2) + (y^2 / b^2) = 1

where:

• a^2 = 49
• b^2 = 36

Finding the Vertices:

• Since a^2 > b^2, the major axis is along the x-axis.
• The vertices are (±a, 0).
• a = √49 = 7
• Vertices: (±7, 0)

Finding the Foci:

• c = √(a^2 – b^2) = √(49 – 36) = √13
• The foci are (±c, 0).
• Foci: (±√13, 0)

Finding the Length of the Major Axis:

• Length of major axis = 2a 
• = 2 * 7 
• = 14

Finding the Length of the Minor Axis:

• Length of minor axis = 2b = 2 * 6 = 12

Finding the Eccentricity:

• Eccentricity (e) = c / a = (√13) / 7

Finding the Length of the Latus Rectum:

Length of latus rectum = 2b^2 / a

 = 2 * 36 / 7 

6.

Ans : 

(x^2 / a^2) + (y^2 / b^2) = 1

where:

• a^2 = 100
• b^2 = 400

Finding the Vertices:

• Since a^2 < b^2, the major axis is along the y-axis.
  
• The vertices are (0, ±b).
  
• b = √400 = 20
  
• Vertices: (0, ±20)
  

Finding the Foci:

• c = √(b^2 – a^2) = √(400 – 100) = √300 = 10√3
  
• The foci are (0, ±c).
  
• Foci: (0, ±10√3)
  

Finding the Length of the Major Axis:

• Length of major axis = 2b = 2 * 20 = 40

Finding the Length of the Minor Axis:

• Length of minor axis = 2a = 2 * 10 = 20

Finding the Eccentricity:

• Eccentricity (e) = c / b = (10√3) / 20 = √3 / 2

Finding the Length of the Latus Rectum:

• Length of latus rectum = 2a^2 / b = 2 * 100 / 20 = 10

7. 36x2 + 4y2 = 144 

Ans : 

Step 1: Convert the equation to standard form.

To put the equation in standard form, divide both sides by 144:

(36x^2 / 144) + (4y^2 / 144) = 144 / 144

Simplify:

x^2 / 4 + y^2 / 36 = 1

Step 2: Identify a^2 and b^2.

From the standard form, we can see that:

• a^2 = 4
• b^2 = 36

Step 3: Calculate c.

The value of c (the distance from the center to the foci) is calculated using the formula:

c = √(b^2 – a^2)

Substituting the values:

c = √(36 – 4) 

= √32 

= 4√2

Step 4: Determine the major and minor axes.

Since a^2 < b^2, the major axis is along the y-axis, and the minor axis is along the x-axis.

• Length of major axis = 2b 
• = 2 * √36 
• = 12
• Length of minor axis = 2a = 2 * √4 = 4

Step 5: Find the coordinates of the foci, vertices, and other points.

• Vertices: (0, ±b) = (0, ±6)
• Foci: (0, ±c) = (0, ±4√2)
• Eccentricity (e): e = c / b = (4√2) / 6 = 2√2 / 3
• Length of latus rectum: 2a^2 / b = 2 * 4 / 6 = 4/3

8. 162+y2=16

Ans : 

Step 1: Convert the equation to standard form.

To put the equation in standard form, divide both sides by 16:

(x^2 / 1) + (y^2 / 16) = 1

Step 2: Identify a^2 and b^2.

• a^2 = 1
• b^2 = 16

Step 3: Calculate c.

The value of c (the distance from the center to the foci) is calculated using the formula:

c = √(b^2 – a^2)

Substituting the values:

c = √(16 – 1) = √15

Step 4: Determine the major and minor axes.

Since a^2 < b^2, the major axis is along the y-axis, and the minor axis is along the x-axis.

• Length of major axis = 2b = 2 * √16 = 8
• Length of minor axis = 2a = 2 * √1 = 2

Step 5: Find the coordinates of the foci, vertices, and other points.

• Vertices: (0, ±b) = (0, ±4)
• Foci: (0, ±c) = (0, ±√15)
• Eccentricity (e): e = c / b = (√15) / 4
• Length of latus rectum: 2a^2 / b 
• = 2 * 1 / 4 
• = 1/2

9. 4x2 + 9y2 = 36

Ans : 

Step 1: Convert the equation to standard form.

To put the equation in standard form, divide both sides by 36:

(4x^2 / 36) + (9y^2 / 36) = 36 / 36

Simplify:

x^2 / 9 + y^2 / 4 = 1

Step 2: Identify a^2 and b^2.

• a^2 = 9
• b^2 = 4

Step 3: Calculate c.

The value of c (the distance from the center to the foci) is calculated using the formula:

c = √(a^2 – b^2)

Substituting the values:

c = √(9 – 4) = √5

Step 4: Determine the major and minor axes.

Since a^2 > b^2, the major axis is along the x-axis, and the minor axis is along the y-axis.

• Length of major axis = 2a = 2 * √9 = 6
• Length of minor axis = 2b = 2 * √4 = 4

Step 5: Find the coordinates of the foci, vertices, and other points.

• Vertices: (±a, 0) = (±3, 0)
• Foci: (±c, 0) = (±√5, 0)
• Eccentricity (e): e = c / a = (√5) / 3
• Length of latus rectum: 2b^2 / 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: 

10. Vertices (± 5, 0), foci (± 4, 0) 

Ans : 

Given:

• Vertices: (±5, 0)
• Foci: (±4, 0)

From the vertices, we can determine:

• a = 5

From the foci, we can determine:

• c = 4

• c^2 = a^2 – b^2
• 16 = 25 – b^2
• b^2 = 9

(x^2 / 25) + (y^2 / 9) = 1

11. Vertices (0, ± 13), foci (0, ± 5)

Ans : 

Given:

• Vertices: (0, ±13)
• Foci: (0, ±5)

From the vertices, we can determine:

• b = 13

From the foci, we can determine:

• c = 5

• c^2 = b^2 – a^2
• 25 = 169 – a^2
• a^2 = 144

(x^2 / 144) + (y^2 / 169) = 1

Therefore, the equation of the ellipse is:

(x^2 / 144) + (y^2 / 169) = 1

12. Vertices (± 6, 0), foci (± 4, 0)

Ans : 

Given:

• Vertices: (±6, 0)
• Foci: (±4, 0)

From the vertices, we can determine:

• a = 6

From the foci, we can determine:

• c = 4

• c^2 = a^2 – b^2
• 16 = 36 – b^2
• b^2 = 20

(x^2 / 36) + (y^2 / 20) = 1

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

Ans :

Given:

• Ends of major axis: (±3, 0)
• Ends of minor axis: (0, ±2)

From the given points, we can determine:

• a = 3
• b = 2

(x^2 / 9) + (y^2 / 4) = 1

14. Ends of major axis (0, ± 5 ), ends of minor axis (± 1, 0)

Ans : 

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

Ans : 

Given:

• Length of major axis = 26
• Foci: (±5, 0)

• 2a = 26
• a = 13

From the foci, we can determine:

• c = 5

• c^2 = a^2 – b^2
• 25 = 169 – b^2
• b^2 = 144

(x^2 / 169) + (y^2 / 144) = 1

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

Ans : 

17. Foci (± 3, 0), a = 4

Ans : 

Given:

• Foci: (±3, 0)
• a = 4

From the foci, we can determine:

• c = 3  

• c^2 = a^2 – b^2
• 9 = 16 – b^2
• b^2 = 7  

(x^2 / 16) + (y^2 / 7) = 1

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

Ans : 

Given:

• b = 3
• c = 4
• Center: (0, 0)

Since the center is at the origin, the standard form remains the same.

• c^2 = a^2 – b^2
• 16 = a^2 – 9
• a^2 = 25

(x^2 / 25) + (y^2 / 9) = 1

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

Ans : 

20. Major axis on the x-axis and passes through the points (4,3) and (6,2).

Ans :