Chapter 11: Conic Sections Exercise 11.3

Given:

Vertices (0, ± 13) and foci (0, ± 5)

Here, the vertices are on the y-axis.

So, the equation of the ellipse will be of the form x²/b² + y²/a² = 1, where ‘a’ is the semi-major axis.

Then, a =13 and c = 5.

It is known that a² = b² + c².

13² = b²+5²

169 = b² + 15

b² = 169 – 125

b = √144

= 12

∴ The equation of the ellipse is x²/12² + y²/13² = 1 or x²/144 + y²/169 = 1

Given:

Vertices (± 6, 0) and foci (± 4, 0)

Here, the vertices are on the x-axis.

So, the equation of the ellipse will be of the form x²/a² + y²/b² = 1, where ‘a’ is the semi-major axis.

Then, a = 6 and c = 4.

It is known that a² = b² + c².

6² = b²+4²

36 = b² + 16

b² = 36 – 16

b = √20

∴ The equation of the ellipse is x²/6² + y²/(√20)² = 1 or x²/36 + y²/20 = 1

Given:

Ends of the major axis (± 3, 0) and ends of minor axis (0, ±2)

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

So, the equation of the ellipse will be of the form x²/a² + y²/b² = 1, where ‘a’ is the semi-major axis.

Then, a = 3 and b = 2.

∴ The equation for the ellipse x²/3² + y²/2² = 1 or x²/9 + y²/4 = 1

Given:

Ends of major axis (0, ±√5) and ends of minor axis (±1, 0)

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

So, the equation of the ellipse will be of the form x²/b² + y²/a² = 1, where ‘a’ is the semi-major axis.

Then, a = √5 and b = 1.

∴ The equation for the ellipse x²/1² + y²/(√5)² = 1 or x²/1 + y²/5 = 1

Given:

Length of the major axis is 26 and the foci (±5, 0)

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

So, the equation of the ellipse will be of the form x²/a² + y²/b² = 1, where ‘a’ is the semi-major axis.

Then, 2a = 26

a = 13 and c = 5.

It is known that a² = b² + c².

13² = b²+5²

169 = b² + 25

b² = 169 – 25

b = √144

= 12

∴ The equation of the ellipse is x²/13² + y²/12² = 1 or x²/169 + y²/144 = 1

Given:

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

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

So, the equation of the ellipse will be of the form x²/b² + y²/a² = 1, where ‘a’ is the semi-major axis.

Then, 2b =16

b = 8 and c = 6.

It is known that a² = b² + c².

a² = 8² + 6²

= 64 + 36

=100

a = √100

= 10

∴ The equation of the ellipse is x²/8² + y²/10² =1 or x²/64 + y²/100 = 1

Given:

Foci (±3, 0) and a = 4

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

So, the equation of the ellipse will be of the form x²/a² + y²/b² = 1, where ‘a’ is the semi-major axis.

Then, c = 3 and a = 4.

It is known that a² = b² + c².

a² = 8² + 6²

= 64 + 36

= 100

16 = b² + 9

b² = 16 – 9

= 7

∴ The equation of the ellipse is x²/16 + y²/7 = 1

Given:

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

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

So, the equation of the ellipse will be of the form x²/a² + y²/b² = 1, where ‘a’ is the semi-major axis.

Then, b = 3 and c = 4.

It is known that a² = b² + c².

a² = 3² + 4²

= 9 + 16

=25

a = √25

= 5

∴ The equation of the ellipse is x²/5² + y²/3² or x²/25 + y²/9 = 1

Given:

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

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/a² = 1, where ‘a’ is the semi-major axis.

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

So, by putting the values x = 3 and y = 2, we get,

3²/b² + 2²/a² = 1

9/b² + 4/a²…. (1)

And by putting the values x = 1 and y = 6, we get,

1¹/b² + 6²/a² = 1

1/b² + 36/a² = 1 …. (2)

On solving equations (1) and (2), we get

b² = 10 and a² = 40.

∴ The equation of the ellipse is x²/10 + y²/40 = 1 or 4x² + y ² = 40

Given:

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

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

So by putting the values x = 4 and y = 3 in equation (1), we get,

16/a² + 9/b² = 1 …. (2)

Putting x = 6 and y = 2 in equation (1), we get,

36/a² + 4/b² = 1 …. (3)

From equation (2)

16/a² = 1 – 9/b²

1/a² = (1/16 (1 – 9/b²)) …. (4)

Substituting the value of 1/a² in equation (3), we get,

36/a² + 4/b² = 1

36(1/a²) + 4/b² = 1

36[1/16 (1 – 9/b²)] + 4/b² = 1

36/16 (1 – 9/b²) + 4/b² = 1

9/4 (1 – 9/b²) + 4/b² = 1

9/4 – 81/4b² + 4/b² = 1

-81/4b² + 4/b² = 1 – 9/4

(-81+16)/4b² = (4-9)/4

-65/4b² = -5/4

-5/4(13/b²) = -5/4

13/b² = 1

1/b² = 1/13

b² = 13

Now substitute the value of b² in equation (4) we get,

1/a² = 1/16(1 – 9/b²)

= 1/16(1 – 9/13)

= 1/16((13-9)/13)

= 1/16(4/13)

= 1/52

a² = 52

Equation of ellipse is x²/a² + y²/b² = 1

By substituting the values of a² and b² in the above equation, we get,

x²/52 + y²/13 = 1

Exercise 11.1 Solutions 15 Questions

Exercise 11.2 Solutions 12 Questions

Exercise 11.4 Solutions 15 Questions

Miscellaneous Exercise On Chapter 11 Solutions 8 Questions

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