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Chapter 11 Conic Sections Ex-11.3 Interview Questions Answers

Question 1 : 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.

x2/36 + y2/16= 1

Answer 1 :

Given:

The equation is x2/36+ y2/16 = 1

Here, the denominator of x2/36 isgreater than the denominator of y2/16.

So, the major axis is along the x-axis, while the minor axis isalong the y-axis.

On comparing the given equation with x2/a2 +y2/b2 =1, we get

a = 6 and b = 4.

c = √(a2 – b2)

= √(36-16)

= √20

= 2√5

Then,

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 = 2 (6) = 12

Length of minor axis = 2b = 2 (4) = 8

Eccentricity, e = c/a =2√5/6 = √5/3

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

Question 2 : 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.

x2/4 + y2/25= 1

Answer 2 :

Given:

The equation is x2/4 +y2/25 = 1

Here, the denominator of y2/25 isgreater than the denominator of x2/4.

So, the major axis is along the x-axis, while the minor axis isalong the y-axis.

On comparing the given equation with x2/a2 + y2/b2 = 1, we get

a = 5 and b = 2.

c = √(a2 – b2)

= √(25-4)

= √21

Then,

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 = 2 (5) = 10

Length of minor axis = 2b = 2 (2) = 4

Eccentricity, e = c/a = √21/5

Length of latus rectum = 2b2/a= (2×22)/5 = (2×4)/5 = 8/5

Question 3 : 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.

x2/16 + y2/9= 1

Answer 3 :

Given:

The equation is x2/16+ y2/9 = 1 or x2/42 + y2/32 = 1

Here, the denominator of x2/16 isgreater than the denominator of y2/9.

So, the major axis is along the x-axis, while the minor axis isalong the y-axis.

On comparing the given equation with x2/a2 +y2/b2 =1, we get

a = 4 and b = 3.

c = √(a2 – b2)

= √(16-9)

= √7

Then,

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 = 2 (4) = 8

Length of minor axis = 2b = 2 (3) = 6

Eccentricity, e = c/a = √7/4

Length of latus rectum = 2b2/a= (2×32)/4 = (2×9)/4 = 18/4 = 9/2

Question 4 : 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.

x2/25 + y2/100= 1

Answer 4 :

Given:

The equation is x2/25+ y2/100 = 1

Here, the denominator of y2/100 isgreater than the denominator of x2/25.

So, the major axis is along the y-axis, while the minor axis isalong the x-axis.

On comparing the given equation with x2/b2 +y2/a2 =1, we get

b = 5 and a =10.

c = √(a2 – b2)

= √(100-25)

= √75

= 5√3

Then,

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 = 2 (10) = 20

Length of minor axis = 2b = 2 (5) = 10

Eccentricity, e = c/a = 5√3/10 = √3/2

Length of latus rectum = 2b2/a= (2×52)/10 = (2×25)/10 = 5

Question 5 : 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.

x2/49 + y2/36= 1

Answer 5 :

Given:

The equation is x2/49+ y2/36 = 1

Here, the denominator of x2/49 isgreater than the denominator of y2/36.

So, the major axis is along the x-axis, while the minor axis isalong the y-axis.

On comparing the given equation with x2/a2 +y2/b2 =1, we get

b = 6 and a =7

c = √(a2 – b2)

= √(49-36)

= √13

Then,

The coordinates of the foci are (√13, 0) and (-√3, 0).

The coordinates of the vertices are (7, 0) and (-7, 0)

Length of major axis = 2a = 2 (7) = 14

Length of minor axis = 2b = 2 (6) = 12

Eccentricity, e = c/a = √13/7

Length of latus rectum = 2b2/a= (2×62)/7 = (2×36)/7 = 72/7

Question 6 : 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.

x2/100 + y2/400= 1

Answer 6 :

Given:

The equation is x2/100+ y2/400 = 1

Here, the denominator of y2/400 isgreater than the denominator of x2/100.

So, the major axis is along the y-axis, while the minor axis isalong the x-axis.

On comparing the given equation with x2/b2 +y2/a2 =1, we get

b = 10 and a =20.

c = √(a2 – b2)

= √(400-100)

= √300

= 10√3

Then,

The coordinates of the foci are (0, 10√3) and (0, -10√3).

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

Length of major axis = 2a = 2 (20) = 40

Length of minor axis = 2b = 2 (10) = 20

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

Length of latus rectum = 2b2/a= (2×102)/20 = (2×100)/20 = 10

Question 7 : 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.

36x2 + 4y2 =144

Answer 7 :

Given:

The equation is 36x2 +4y2 = 144 or x2/4 + y2/36= 1 or x2/22 +y2/62 =1

Here, the denominator of y2/62 is greater than the denominator of x2/22.

So, the major axis is along the y-axis, while the minor axis isalong the x-axis.

On comparing the given equation with x2/b2 +y2/a2 =1, we get

b = 2 and a = 6.

c = √(a2 – b2)

= √(36-4)

= √32

= 4√2

Then,

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

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

Length of major axis = 2a = 2 (6) = 12

Length of minor axis = 2b = 2 (2) = 4

Eccentricity, e = c/a = 4√2/6 = 2√2/3

Length of latus rectum = 2b2/a= (2×22)/6 = (2×4)/6 = 4/3

Question 8 : 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.

16x2 + y2 =16

Answer 8 :

Given:

The equation is 16x2 +y2 = 16 or x2/1 + y2/16= 1 or x2/12 +y2/42 =1

Here, the denominator of y2/42 is greater than the denominator of x2/12.

So, the major axis is along the y-axis, while the minor axis isalong the x-axis.

On comparing the given equation with x2/b2 +y2/a2 =1, we get

b =1 and a =4.

c = √(a2 – b2)

= √(16-1)

= √15

Then,

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

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

Length of major axis = 2a = 2 (4) = 8

Length of minor axis = 2b = 2 (1) = 2

Eccentricity, e = c/a = √15/4

Length of latus rectum = 2b2/a= (2×12)/4 = 2/4 = ½

Question 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.

4x2 + 9y2 =36

Answer 9 :

Given:

The equation is 4x2 +9y2 = 36 or x2/9 + y2/4= 1 or x2/32 +y2/22 =1

Here, the denominator of x2/32 is greater than the denominator of y2/22.

So, the major axis is along the x-axis, while the minor axis isalong the y-axis.

On comparing the given equation with x2/a2 +y2/b2 =1, we get

a =3 and b =2.

c = √(a2 – b2)

= √(9-4)

= √5

Then,

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 = 2 (3) = 6

Length of minor axis = 2b = 2 (2) = 4

Eccentricity, e = c/a = √5/3

Length of latus rectum = 2b2/a= (2×22)/3 = (2×4)/3 = 8/3

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

Answer 10 :

Given:

Vertices (± 5, 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 = 5 and c = 4.

It is known that a2 = b+ c2.

So, 52 = b+ 42

25 = b2 + 16

b2 = 25 – 16

b = √9

= 3

Theequation of the ellipse is x2/52 + y2/32 =1 or x2/25 + y2/9 = 1


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Chapter 11 Conic Sections Ex-11.3 Contributors

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