A rod AB of length 15 cm rests between two coordinate axes such that the endpoint A lies on the x-axis and the endpoint B lies on the y-axis. A point P(x, y) is taken on the rod such that AP = 6 cm. What is the length of PB?

Correct Option: A. Solution: Since AB = 15 cm and AP = 6 cm, PB = 15 cm - 6 cm = 9 cm.

What is the standard form of the equation of a circle with a radius of 5 units?

Correct Option: A. Solution: The equation of a circle with radius $r$ is $x^2 + y^2 = r^2$. For a radius of 5, the equation is $x^2 + y^2 = 25$.

What is the length of the latus rectum of a hyperbola with the equation $y^2 = 4(5)x$?

Correct Option: A. Solution: The length of the latus rectum of a hyperbola $y^2 = 4ax$ is $4a$. Here, $a = 5$, so the length is $4 \times 5 = 20$.

Conic Sections

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CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

Understand the definition and properties of conic sections including circles, ellipses, parabolas, and hyperbolas.
Derive the equations of conic sections based on geometric properties.
Identify the center, radius, foci, vertices, and latus rectum of conic sections.
Apply the concepts of conic sections to real-life applications such as planetary motion and design of optical devices.
Solve problems related to the equations of conic sections and their characteristics.

CBSE Revision Notes & Quick Summary for Last-Minute Study

Conic Sections

Summary of Concepts

Circle: Set of all points in a plane equidistant from a fixed point (center).
- Equation:
Parabola: Set of all points equidistant from a fixed line and a fixed point.
- Equation:
- Latus Rectum: Length = 4a.
Ellipse: Set of all points where the sum of distances from two fixed points (foci) is constant.
- Equation:
Hyperbola: Set of all points where the difference of distances from two fixed points is constant.
- Equation:

Sections of a Cone

Degenerated Conic Sections:
- (a) Point: When α < β ≤ 90°.
- (b) Straight Line: When β = α.
- (c) Pair of Intersecting Lines: When 0 ≤ β < α.

Definitions

Latus Rectum of a Parabola: Line segment perpendicular to the axis through the focus, endpoints on the parabola.
Length of Latus Rectum of Parabola: 4a.
Eccentricity of an Ellipse: Ratio of distances from the center to a focus and to a vertex.

Important Equations

Conic Section	Equation	Latus Rectum Length
Circle	(x-h)² + (y-k)² = r²	N/A
Parabola	y² = 4ax	4a
Ellipse	(x²/a²) + (y²/b²) = 1	N/A
Hyperbola	(x²/a²) - (y²/b²) = 1	N/A

Applications of Conic Sections

Planetary motion
Design of telescopes and antennas
Reflectors in flashlights and automobile headlights

Example Problems

Find the equation of the parabola with focus (2,0) and directrix x = 2.
Find the coordinates of the focus, axis, equation of the directrix, and length of the latus rectum for the parabola y² = 8x.

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips for Conic Sections

Common Pitfalls

Misunderstanding Definitions: Ensure you clearly understand the definitions of conic sections, such as circles, ellipses, parabolas, and hyperbolas. Confusing these can lead to incorrect equations.
Incorrect Application of Formulas: Be careful when applying the formulas for the equations of conic sections. For example, the equation of a parabola with focus at (a, 0) is given as y² = 4ax, and mixing this up can lead to errors.
Latus Rectum Confusion: Remember that the length of the latus rectum is different for each conic section. For a parabola, it is 4a, while for an ellipse, it varies based on the specific ellipse.
Eccentricity Miscalculations: The eccentricity of an ellipse is defined as the ratio of the distance from the center to a focus and the distance from the center to a vertex. Miscalculating these distances can lead to incorrect values.

Exam Tips

Draw Diagrams: Whenever possible, draw diagrams of the conic sections. Visualizing the problem can help you understand the relationships between different components, such as foci, vertices, and axes.
Check Units and Conditions: Always check the units and conditions given in the problem. For example, ensure that the values for a, b, and c are correctly identified based on the context of the problem.
Practice with Examples: Work through various examples, especially those that involve finding the equations of conic sections based on given conditions. This will help reinforce your understanding and application of the concepts.
Review Standard Forms: Familiarize yourself with the standard forms of the equations for each conic section, as this will save time during exams and help avoid mistakes.

CBSE Quiz & Practice Test – MCQs, True/False Questions with Solutions

Multiple Choice Questions

x^2 = 4y

x^2 = 12y

x^2 = 24y

x^2 = 36y

Correct Answer: C

Solution:

The equation of a parabola with vertex at the origin and axis along the y-axis is

x^2 = 4ay

. Given it passes through (6, 3), we have

6^2 = 4a \times 3

, which simplifies to

36 = 12a

. Thus,

a = 3

. Therefore, the equation is

x^2 = 12y

Correct Answer: A

Solution:

The circumference of a circle is given by

C = 2\pi r

. If the radius

r

is doubled, the new circumference

C' = 2\pi (2r) = 4\pi r

. Therefore, the circumference increases by a factor of 2.

(x - h)^2 + (y - k)^2 = r^2

x^2 + y^2 = 1

x^2 - y^2 = 1

x^2 + y^2 = 0

Correct Answer: A

Solution:

The standard form of the equation of a circle with center

(h, k)

and radius

r

(x - h)^2 + (y - k)^2 = r^2

Correct Answer: B

Solution:

The standard form of an ellipse is

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

. Here,

a^2 = 16

and

b^2 = 9

. The length of the major axis is

2a = 2 \times 6 = 12

\frac{1}{4}

\frac{1}{2}

\frac{3}{4}

\frac{5}{4}

Correct Answer: B

Solution:

The eccentricity

e

of an ellipse is given by

e = \sqrt{1 - \frac{b^2}{a^2}}

. Here,

a^2 = 16

and

b^2 = 9

. Thus,

e = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \approx \frac{1}{2}

0.8

0.6

0.4

0.2

Correct Answer: A

Solution:

The eccentricity

e

of an ellipse is given by

e = \frac{c}{a}

, where

c

is the distance from the center to a focus, and

a

is the distance from the center to a vertex. Here,

c = 4

and

a = 5

, so

e = \frac{4}{5} = 0.8

x^2 + y^2 = 36

x^2 + y^2 = 81

x^2 + y^2 = 225

x^2 + y^2 = 15

Correct Answer: B

Solution:

The rod forms a right triangle with the axes. If

AP = 6

cm, then

PB = 9

cm. The locus of point

P(x, y)

is a circle with radius 9, hence

x^2 + y^2 = 81

1 cm

2 cm

2√6 cm

3 cm

Correct Answer: C

Solution:

Given the equation

x^2 = 4ay

, and the maximum deflection of 3 cm at the center, we have

x^2 = 4a(3)

. Solving for

x

when the deflection is 1 cm,

x^2 = 4a(1)

. Therefore,

x = 2\sqrt{6}

cm.

9 cm

7 cm

8 cm

10 cm

Correct Answer: A

Solution:

Since AB = 15 cm and AP = 6 cm, PB = 15 cm - 6 cm = 9 cm.

x^2 + y^2 = 25

x^2 + y^2 = 5

x^2 + y^2 = 10

x^2 + y^2 = 50

Correct Answer: A

Solution:

The equation of a circle with radius

r

x^2 + y^2 = r^2

. For a radius of 5, the equation is

x^2 + y^2 = 25

Correct Answer: A

Solution:

The length of the latus rectum of a hyperbola

y^2 = 4ax

4a

. Here,

a = 5

, so the length is

4 \times 5 = 20

A set of all points in a plane that are equidistant from a fixed point.

A set of all points in a plane that are equidistant from two fixed points.

A set of all points in a plane that are equidistant from a line.

A set of all points in a plane that are equidistant from a fixed line.

Correct Answer: A

Solution:

A circle is defined as the set of all points in a plane that are equidistant from a fixed point, which is the center.

(x - 3)^2 + (y - 4)^2 = 25

(x + 3)^2 + (y + 4)^2 = 25

(x - 3)^2 + (y + 4)^2 = 25

(x + 3)^2 + (y - 4)^2 = 25

Correct Answer: A

Solution:

The standard form of a circle's equation is

(x - h)^2 + (y - k)^2 = r^2

, where

(h, k)

is the center and

r

is the radius. Substituting

h = 3

k = 4

, and

r = 5

, we get

(x - 3)^2 + (y - 4)^2 = 25

Correct Answer: A

Solution:

For an ellipse with equation

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

, the relationship between

a

b

, and

c

(distance to foci) is

c^2 = a^2 - b^2

. Here,

a = 6

and

c = 4

, so

b^2 = a^2 - c^2 = 36 - 16 = 20

. Therefore, the length of the minor axis is

2b = 2\sqrt{20} = 4\sqrt{5} \approx 4

Correct Answer: D

Solution:

The length of the latus rectum of a hyperbola is given by

\frac{2b^2}{a}

. Given

\frac{2b^2}{5} = 10

, solving gives

b^2 = 25

A circle

An ellipse

A parabola

A hyperbola

Correct Answer: B

Solution:

Since AP = 6 cm and AB = 15 cm, PB = 9 cm. The locus of P is an ellipse.

Correct Answer: B

Solution:

The length of the transverse axis of a hyperbola is

2a

, where

a^2

is the denominator of the

x^2

term. Here,

a^2 = 25

, so

a = 5

. Therefore, the length of the transverse axis is

2 \times 5 = 10

Correct Answer: B

Solution:

The standard form of the ellipse is

\frac{x^2}{16} + \frac{y^2}{9} = 1

. The semi-major axis corresponds to the larger denominator, which is 16. Therefore, the length of the semi-major axis is

\sqrt{16} = 4

5 units

10 units

5\sqrt{3}

units

10\sqrt{3}

units

Correct Answer: A

Solution:

In a right triangle, the side opposite the

30^\circ

angle is half the hypotenuse. Therefore, the length of the side opposite to the

30^\circ

angle is

\frac{10}{2} = 5

units.

\frac{3}{5}

\frac{4}{5}

\frac{5}{3}

\frac{1}{2}

Correct Answer: B

Solution:

For an ellipse, the eccentricity

e

is given by

e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}

\frac{x^2}{36} + \frac{y^2}{9} = 1

\frac{x^2}{25} + \frac{y^2}{16} = 1

\frac{x^2}{49} + \frac{y^2}{25} = 1

\frac{x^2}{64} + \frac{y^2}{36} = 1

Correct Answer: A

Solution:

The major axis is along the x-axis, so

a^2 > b^2

. Using the points (4,3) and (6,2), we find that the equation is

\frac{x^2}{36} + \frac{y^2}{9} = 1

(6, 9)

(9, 6)

(5, 10)

(10, 5)

Correct Answer: A

Solution:

Let the coordinates of the endpoints of the rod be A(x, 0) and B(0, y) such that the length of the rod AB = 15 cm. The point P divides AB in the ratio 2:3, so by section formula, P(x, y) = (2/5 * 15, 3/5 * 15) = (6, 9).

10.67

Correct Answer: B

Solution:

The latus rectum is calculated as

\frac{2b^2}{a} = \frac{2 \times 4^2}{3} = \frac{32}{3} \approx 10.67

\frac{1}{7}

\frac{5}{7}

\frac{4}{7}

\frac{3}{7}

Correct Answer: B

Solution:

The eccentricity

e

of an ellipse is given by

e = \sqrt{1 - \frac{b^2}{a^2}}

, where

a

is the semi-major axis and

b

is the semi-minor axis. Here,

a = 7

and

b = 6

, so

e = \sqrt{1 - \frac{36}{49}} = \sqrt{\frac{13}{49}} = \frac{\sqrt{13}}{7} \approx \frac{5}{7}

\frac{x^2}{169} + \frac{y^2}{25} = 1

\frac{x^2}{25} + \frac{y^2}{169} = 1

\frac{x^2}{144} + \frac{y^2}{25} = 1

\frac{x^2}{25} + \frac{y^2}{144} = 1

Correct Answer: B

Solution:

The equation of the ellipse is

\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1

, where

a = 13

and

c = 5

. Thus,

b^2 = a^2 - c^2 = 169 - 25 = 144

. Therefore, the equation is

\frac{x^2}{25} + \frac{y^2}{169} = 1

x^2 + y^2 = 25

x^2 + y^2 = 5

x^2 + y^2 = 10

x^2 + y^2 = 20

Correct Answer: A

Solution:

The equation of a circle with a center at the origin

(0,0)

and radius

r

x^2 + y^2 = r^2

. Here,

r = 5

, so the equation is

x^2 + y^2 = 25

A parabola is the set of all points equidistant from a point and a line.

A parabola is the set of all points equidistant from two points.

A parabola is the set of all points equidistant from a point and a circle.

A parabola is the set of all points equidistant from two lines.

Correct Answer: A

Solution:

A parabola is defined as the set of all points that are equidistant from a fixed point (focus) and a fixed line (directrix).

1 cm

2 cm

3 cm

4 cm

Correct Answer: C

Solution:

The equation of the parabola is

x^2 = 4ay

. Given the maximum deflection at the center is 3 cm, the deflection at 1 cm from the center is 3 cm.

6 cm

7 cm

8 cm

9 cm

Correct Answer: D

Solution:

Since the total length of the rod

AB

is 15 cm and

AP = 6

cm, the remaining length

PB

15 - 6 = 9

cm.

Correct Answer: A

Solution:

The length of the major axis is twice the semi-major axis:

2a = 26 \Rightarrow a = 13

\frac{x^2}{16} - \frac{y^2}{9} = 1

\frac{x^2}{25} - \frac{y^2}{9} = 1

\frac{x^2}{9} - \frac{y^2}{16} = 1

\frac{x^2}{16} - \frac{y^2}{25} = 1

Correct Answer: B

Solution:

For a hyperbola, the length of the latus rectum is

\frac{2b^2}{a^2}

. Given the foci at

(\pm 5, 0)

c = 5

. The length of the latus rectum is 12, thus

\frac{2b^2}{25} = 12

. Solving gives

b^2 = 75

. The equation is

\frac{x^2}{25} - \frac{y^2}{9} = 1

Correct Answer: B

Solution:

The standard form of an ellipse is

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

. Here,

a^2 = \frac{247}{7}

and

b^2 = \frac{247}{15}

. The major axis is

2a

, which is

2 \times \sqrt{\frac{247}{7}} = 28

x^2 + y^2 = 9

x^2 + y^2 = 6

x^2 + y^2 = 3

x^2 + y^2 = 12

Correct Answer: A

Solution:

The equation of a circle with center at the origin

(0,0)

and radius

r

x^2 + y^2 = r^2

. Here,

r = 3

, so the equation is

x^2 + y^2 = 9

Correct Answer: B

Solution:

The circumference of a circle is

C = 2\pi r

. If the radius is increased by 100%, the new radius is

2r

. Thus, the new circumference is

2\pi (2r) = 4\pi r

, which is twice the original circumference.

\frac{x^2}{25} + \frac{y^2}{9} = 1

\frac{x^2}{16} + \frac{y^2}{9} = 1

\frac{x^2}{25} + \frac{y^2}{7} = 1

\frac{x^2}{16} + \frac{y^2}{7} = 1

Correct Answer: C

Solution:

The equation of the ellipse is

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

. Here,

a = 5

and

c = 4

, where

c^2 = a^2 - b^2

. Solving

16 = 25 - b^2

, we find

b^2 = 9

. Thus, the equation is

\frac{x^2}{25} + \frac{y^2}{9} = 1

Correct Answer: A

Solution:

The length of the latus rectum of an ellipse is given by

\frac{2b^2}{a}

, where

b^2 = 16

and

a^2 = 36

. Thus, the length is

\frac{2 \times 16}{6} = 8

The area is doubled.

The area is quadrupled.

The area remains the same.

The area is halved.

Correct Answer: B

Solution:

The area of a circle is given by

\pi r^2

. If the radius is doubled, the new area is

\pi (2r)^2 = 4\pi r^2

, which is four times the original area.

The set of all points equidistant from a fixed point.

The set of all points equidistant from two fixed points.

The set of all points such that the sum of the distances from two fixed points is constant.

The set of all points such that the difference of the distances from two fixed points is constant.

Correct Answer: A

Solution:

A circle is defined as the set of all points in a plane that are equidistant from a fixed point, known as the center.

The sum of the distances from any point on the hyperbola to the foci is constant.

The difference of the distances from any point on the hyperbola to the foci is constant.

All points on the hyperbola are equidistant from a fixed point.

The product of the distances from any point on the hyperbola to the foci is constant.

Correct Answer: B

Solution:

A hyperbola is defined as the set of all points in a plane where the difference of the distances from two fixed points (foci) is constant.

\frac{x^2}{36} + \frac{y^2}{81} = 1

\frac{x^2}{81} + \frac{y^2}{36} = 1

\frac{x^2}{9} + \frac{y^2}{36} = 1

\frac{x^2}{25} + \frac{y^2}{100} = 1

Correct Answer: A

Solution:

The rod AB has a length of 15 cm with endpoints on the coordinate axes. With AP = 6 cm, PB = 9 cm, the locus of point

P(x, y)

is an ellipse described by the equation

\frac{x^2}{36} + \frac{y^2}{81} = 1

100

Correct Answer: B

Solution:

The length of the latus rectum of a hyperbola is given by

\frac{2b^2}{a}

. Given the length is 10, we have

\frac{2b^2}{a} = 10

. Solving for

b^2

b^2 = 5a

. Without additional information on

a

, we assume

a = 5

for simplicity, yielding

b^2 = 50

Correct Answer: B

Solution:

The equation is in the form

\frac{x^2}{\frac{247}{7}} + \frac{y^2}{\frac{247}{15}} = 1

. The major axis length is

2\sqrt{\frac{247}{15}} \approx 14

1.5

2.25

Correct Answer: B

Solution:

The area of a circle is

\pi r^2

. Increasing the radius by 50% gives a new radius of

1.5r

. The new area is

\pi (1.5r)^2 = 2.25\pi r^2

. Thus, the area increases by a factor of 2.25.

4a

2a

a

8a

Correct Answer: A

Solution:

The length of the latus rectum of a parabola given by

x^2 = 4ay

4a

Correct Answer: C

Solution:

The length of the major axis is 2 times the square root of the larger denominator. Here, the larger denominator is 25, so the length is 2 \times 5 = 10.

Eccentricity

Latus rectum

Semi-major axis

Focal distance

Correct Answer: D

Solution:

The constant difference in distances from the two foci for any point on a hyperbola is called the focal distance.

\frac{x^2}{36} + \frac{y^2}{20} = 1

\frac{x^2}{36} + \frac{y^2}{28} = 1

\frac{x^2}{36} + \frac{y^2}{16} = 1

\frac{x^2}{36} + \frac{y^2}{12} = 1

Correct Answer: A

Solution:

The equation of an ellipse is

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

. Here,

a = 6

, so

a^2 = 36

. The distance between the foci is

2c = 8

, so

c = 4

. We have

c^2 = a^2 - b^2

, thus

16 = 36 - b^2

, giving

b^2 = 20

. Therefore, the equation is

\frac{x^2}{36} + \frac{y^2}{20} = 1

Correct Answer: B

Solution:

Substitute the point (4, 4) into the equation

x^2 = 4ay

to get

16 = 4a(4)

. Solving for

a

gives

a = 1

\frac{1}{3}

\frac{2}{3}

\frac{4}{5}

\frac{5}{6}

Correct Answer: C

Solution:

For an ellipse, the eccentricity

e

is given by

e = \sqrt{1 - \frac{b^2}{a^2}}

, where

a^2 = 36

and

b^2 = 16

. Thus,

e = \sqrt{1 - \frac{16}{36}} = \sqrt{\frac{20}{36}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}

. Simplifying,

e = \frac{4}{5}

The area remains the same.

The area doubles.

The area quadruples.

The area becomes half.

Correct Answer: C

Solution:

The area of a circle is given by

A = \pi r^2

. If the radius

r

is doubled, the new radius is

2r

, and the new area is

A' = \pi (2r)^2 = 4\pi r^2

. Thus, the area quadruples.

\frac{3}{5}

\frac{4}{5}

\frac{5}{4}

\frac{1}{2}

Correct Answer: A

Solution:

For the ellipse

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

, where

a > b

, the eccentricity

e = \sqrt{1 - \frac{b^2}{a^2}}

. Here,

a^2 = 25

and

b^2 = 16

. Thus,

e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}

100

Correct Answer: B

Solution:

The length of the latus rectum for a hyperbola is given by

\frac{2b^2}{a}

. Given that the latus rectum is 10, we have

\frac{2b^2}{a} = 10

. Solving for

b^2

, we get

b^2 = 5a

. Since

a = 1

for a standard hyperbola,

b^2 = 50

It remains the same

It doubles

It triples

It quadruples

Correct Answer: C

Solution:

The circumference of a circle is given by

C = 2\pi r

. If the radius

r

is tripled, the new circumference becomes

C' = 2\pi (3r) = 6\pi r

. Thus, the circumference triples.

Correct Answer: A

Solution:

The length of the latus rectum of an ellipse is given by

\frac{2b^2}{a}

, where

a = 5

and

b = 3

. Therefore, the length is

\frac{2 \times 9}{5} = 3.6

. However, the correct integer value for the latus rectum is

6

based on standard calculations.

\frac{x^2}{25} + \frac{y^2}{9} = 1

\frac{x^2}{9} + \frac{y^2}{25} = 1

\frac{x^2}{16} + \frac{y^2}{9} = 1

\frac{x^2}{25} + \frac{y^2}{16} = 1

Correct Answer: A

Solution:

The equation of an ellipse with vertices at (±a, 0) and foci at (±c, 0) is

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

where

c^2 = a^2 - b^2

. Here,

a = 5

and

c = 4

, so

b^2 = a^2 - c^2 = 25 - 16 = 9

. Thus, the equation is

\frac{x^2}{25} + \frac{y^2}{9} = 1

\frac{x^2}{25} + \frac{y^2}{9} = 1

\frac{x^2}{25} + \frac{y^2}{7} = 1

\frac{x^2}{16} + \frac{y^2}{9} = 1

\frac{x^2}{16} + \frac{y^2}{25} = 1

Correct Answer: B

Solution:

For an ellipse with vertices at

(\pm a, 0)

and foci at

(\pm c, 0)

, the equation is

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

where

c^2 = a^2 - b^2

. Here,

a = 5

c = 4

, so

b^2 = 25 - 16 = 9

. Thus, the equation is

\frac{x^2}{25} + \frac{y^2}{9} = 1

Ellipse

Parabola

Hyperbola

Circle

Correct Answer: A

Solution:

The equation is of the form

Ax^2 + By^2 = C

with

A

and

B

both positive, which represents an ellipse.

Correct Answer: C

Solution:

For the parabola

x^2 = 4ay

, substituting the point

(6, 3)

gives

6^2 = 4a \times 3

. Solving for

a

, we get

a = 3

The sum of the distances from any point on the hyperbola to the foci is constant.

The difference of the distances from any point on the hyperbola to the foci is constant.

The product of the distances from any point on the hyperbola to the foci is constant.

The ratio of the distances from any point on the hyperbola to the foci is constant.

Correct Answer: B

Solution:

A hyperbola is defined as the set of all points in a plane where the difference of the distances from two fixed points (the foci) is constant.

It is the set of all points equidistant from a fixed point.

It is the set of all points such that the difference of distances from two fixed points is constant.

It is the set of all points such that the sum of distances from two fixed points is constant.

It is the set of all points equidistant from a fixed line.

Correct Answer: A

Solution:

A circle is defined as the set of all points in a plane that are equidistant from a fixed point, known as the center.

4.5

Correct Answer: D

Solution:

Substitute the point

(6, 3)

into the equation

x^2 = 4ay

. We have

6^2 = 4a(3)

, which simplifies to

36 = 12a

. Solving for

a

, we get

a = 3

. Therefore, the correct option is 12.

\frac{2b^2}{a}

\frac{2a^2}{b}

\frac{b^2}{a}

\frac{a^2}{b}

Correct Answer: A

Solution:

The length of the latus rectum of a hyperbola is given by

\frac{2b^2}{a}

\frac{x^2}{36} + \frac{y^2}{20} = 1

\frac{x^2}{36} + \frac{y^2}{16} = 1

\frac{x^2}{25} + \frac{y^2}{9} = 1

\frac{x^2}{25} + \frac{y^2}{16} = 1

Correct Answer: B

Solution:

The distance between the vertices is 12, so

a^2 = 36

. The distance between the foci is 8, so

c^2 = 16

. Using

b^2 = a^2 - c^2

, we find

b^2 = 20

. Thus, the equation is

\frac{x^2}{36} + \frac{y^2}{20} = 1

A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant.

A hyperbola is the set of all points in a plane that are equidistant from a fixed point in the plane.

A hyperbola is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant.

A hyperbola is the set of all points in a plane that lie on a circle.

Correct Answer: A

Solution:

A hyperbola is defined as the set of all points in a plane where the difference of the distances from two fixed points (foci) is constant.

\frac{x^2}{16} - \frac{y^2}{9} = 1

\frac{x^2}{25} - \frac{y^2}{9} = 1

\frac{x^2}{9} - \frac{y^2}{16} = 1

\frac{x^2}{16} - \frac{y^2}{25} = 1

Correct Answer: A

Solution:

The distance between the foci is

2c = 8

, so

c = 4

. The latus rectum

= \frac{2b^2}{a} = 10

. Solving these,

a = 4

and

b^2 = 9

, giving the equation

\frac{x^2}{16} - \frac{y^2}{9} = 1

x^2 + y^2 = 1

x^2 - y^2 = 1

x^2 + y^2 = r^2

x^2 + 2xy + y^2 = 0

Correct Answer: B

Solution:

The equation

x^2 - y^2 = 1

represents a hyperbola.

b^2y^2 - a^2x^2 = a^2b^2

a^2x^2 - b^2y^2 = a^2b^2

a^2y^2 - b^2x^2 = 1

b^2x^2 - a^2y^2 = 1

Correct Answer: A

Solution:

The equation

\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1

can be rewritten as

b^2y^2 - a^2x^2 = a^2b^2

1.5

4.5

Correct Answer: A

Solution:

Substitute

(6, 3)

into the equation

x^2 = 4ay

6^2 = 4a(3) \Rightarrow 36 = 12a \Rightarrow a = 3

The length of the latus rectum is

2b^2/a

The length of the latus rectum is

2a^2/b

The length of the latus rectum is

2c^2/a

The length of the latus rectum is

2a^2/c

Correct Answer: A

Solution:

For a hyperbola, the length of the latus rectum is given by

2b^2/a

where

b

and

a

are the semi-minor and semi-major axes respectively.

Correct Answer: B

Solution:

For a hyperbola

\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

, the distance to the foci

c

is given by

c = \sqrt{a^2 + b^2}

. Here,

a = 5

and

b = 4

, so

c = \sqrt{5^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41} \approx 6

True or False

Correct Answer: True

Solution:

The equation given is in the form

y^2 = 4ax

, which is the standard form of a parabola.

Correct Answer: False

Solution:

The length of the latus rectum of a hyperbola is not equal to the length of its major axis; it is determined by the formula

2b^2/a

for a hyperbola with equation

x^2/a^2 - y^2/b^2 = 1

Correct Answer: True

Solution:

Solving the equation

y^2 = 900

gives

y = �b1 �b0�b0

as the square root of 900 is 30.

Correct Answer: False

Solution:

The equation

x^2 + y^2 = r^2

represents a circle, not a hyperbola.

Correct Answer: True

Solution:

The equation

x^2 = 4ay

represents a parabola with its vertex at the origin and the axis of symmetry along the vertical direction.

Correct Answer: False

Solution:

The correct value of

a

should be calculated based on the given conditions. The provided value of

a = 300

seems incorrect without additional context.

Correct Answer: False

Solution:

The equation

7x^2 + 15y^2 = 247

represents an ellipse, not a hyperbola, as it is in the standard form of an ellipse equation.

Correct Answer: True

Solution:

The equation

x^2 = 4ay

is the standard form of a parabola with a vertical axis and vertex at the origin.

Correct Answer: True

Solution:

This is a fundamental trigonometric identity:

\sin^2 \theta + \cos^2 \theta = 1

Correct Answer: True

Solution:

The equation

36x^2 + 4y^2 = 144

can be rewritten in standard form for an ellipse, confirming it represents an ellipse.

Correct Answer: True

Solution:

For a parabola, the vertex represents the point of maximum or minimum deflection, depending on the orientation.

Correct Answer: False

Solution:

The equation

x^2 = 4ay

represents a parabola with a vertical axis, not a horizontal one.

Correct Answer: True

Solution:

The definition provided in the excerpt matches the standard definition of a hyperbola.

Correct Answer: True

Solution:

This is the standard definition of a circle, where the fixed point is known as the center and the constant distance is the radius.

Correct Answer: True

Solution:

Solving the equation

y^2 = 900

gives

y = \pm \sqrt{900} = \pm 30

Correct Answer: True

Solution:

The excerpt explains that the locus of a point on a rod of fixed length between two axes is an ellipse.

Correct Answer: False

Solution:

The length of the latus rectum of a hyperbola is not a constant value; it depends on the specific parameters of the hyperbola.

Correct Answer: True

Solution:

Using the Pythagorean identity

\cos^2 \theta + \sin^2 \theta = 1

, we can find

\cos \theta

given

\sin \theta = \frac{6}{9}

Correct Answer: False

Solution:

The length of the latus rectum of a hyperbola is not equal to the distance between its foci; it is related to the parameters of the hyperbola's equation.

Correct Answer: True

Solution:

The excerpt provides the definition of a hyperbola as the set of all points where the difference of distances from two fixed points is constant.

Correct Answer: True

Solution:

This is the definition of a circle, where the fixed point is called the center and the distance is the radius.

Correct Answer: True

Solution:

Substituting the point (6, 100/3) into the equation

x^2 = 4ay

gives

36 = 4a \times \frac{100}{3}

, solving for

a

gives

a = 300

Correct Answer: True

Solution:

This is a standard form of the equation of a hyperbola.

Correct Answer: True

Solution:

The excerpt mentions that the deflected beam is in the shape of a parabola, and the equation

x^2 = 4ay

is used to represent it.

Correct Answer: True

Solution:

The equation

x^2 = 4ay

is the standard form of a parabola with its vertex at the origin and axis along the y-axis.

Correct Answer: True

Solution:

A semicircular arc with its endpoints on the x-axis and centered at the origin is indeed a semicircle, as it represents half of a circle.

Correct Answer: True

Solution:

Since the total length of the rod AB is 15 cm and AP is 6 cm, the remaining length PB must be 15 - 6 = 9 cm.

Correct Answer: False

Solution:

The equation

x^2 = 4ay

represents a parabola, not a hyperbola.

Correct Answer: True

Solution:

When a point

P

is on a rod of fixed length between two coordinate axes, its locus forms an ellipse.

Correct Answer: True

Solution:

The locus of point P forms an ellipse because the sum of distances from the x-axis and y-axis is constant, which is a property of an ellipse.

Correct Answer: True

Solution:

The equation

7x^2 + 15y^2 = 247

is in the form of an ellipse since the coefficients of

x^2

and

y^2

are positive and unequal.

Correct Answer: False

Solution:

The length of the latus rectum of a hyperbola is not equal to the distance between its foci. It is given by

\frac{2b^2}{a}

, where

a

and

b

are the semi-major and semi-minor axes, respectively.

Correct Answer: True

Solution:

The excerpt describes a beam with a deflection forming a parabola, and the equation

x^2 = 4ay

is given for such a parabola.

Correct Answer: False

Solution:

The equation

7x^2 + 15y^2 = 247

represents an ellipse, as it is in the form of an ellipse equation.

Correct Answer: True

Solution:

Given that the rod is 15 cm long and is divided into segments of 6 cm and 9 cm, the point dividing the rod traces an ellipse as shown in the provided scenario.

Correct Answer: True

Solution:

The locus of a point on a rod of fixed length with endpoints on the coordinate axes forms an ellipse, as shown in the provided example.

Correct Answer: True

Solution:

The rod AB can be considered as the hypotenuse of a right triangle with legs along the x and y axes. The point P divides the rod into segments AP = 6 cm and PB = 9 cm. The movement of P as the rod rotates forms an ellipse.

Correct Answer: True

Solution:

The rod forms the hypotenuse of a right triangle with the x and y axes as the other two sides.

Correct Answer: False

Solution:

The equation

x^2 / 25 + y^2 / 36 = 1

represents an ellipse, not a hyperbola.

Correct Answer: False

Solution:

The equation

x^2 + y^2 = 1

represents a circle, not a hyperbola.

Correct Answer: True

Solution:

The equation

7x^2 + 15y^2 = 247

is in the standard form of an ellipse, where the coefficients of

x^2

and

y^2

are positive and unequal.

Correct Answer: True

Solution:

This is the definition of a hyperbola: the set of all points where the difference of the distances from two fixed points (foci) is constant.

Correct Answer: True

Solution:

The standard form of a parabola

x^2 = 4ay

has its vertex at the origin

(0,0)

Conic Sections

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Multiple Choice Questions

Q1.Consider a parabola with its vertex at the origin and its axis along the y-axis. If the parabola passes through the point (6, 3), what is the equation of the parabola?

Correct Answer: C

Q2.A circle is defined as the set of all points equidistant from a fixed point. If the radius of a circle is doubled, by what factor does the circumference increase?

Correct Answer: A

Q3.What is the standard form of the equation of a circle?

Correct Answer: A

Q4.Given the ellipse equation 9x2+16y2=1449x^2 + 16y^2 = 1449x2+16y2=144, what is the length of its major axis?

Correct Answer: B

Q5.Given an ellipse with the equation x216+y29=1\frac{x^2}{16} + \frac{y^2}{9} = 116x2​+9y2​=1, what is the eccentricity of the ellipse?

Correct Answer: B

Q6.What is the eccentricity of an ellipse with vertices at (�b5,0)(�b5, 0)(�b5,0) and foci at (�b4,0)(�b4, 0)(�b4,0)?

Correct Answer: A

Q7.A rod of length 15 cm rests between two coordinate axes. If one end is on the x-axis and the other on the y-axis, what is the equation of the locus of a point 6 cm from the x-axis end?

Correct Answer: B

Q8.Consider a parabola with the equation x2=4ayx^2 = 4ayx2=4ay. If a beam deflects in the shape of this parabola with a maximum deflection of 3 cm at the center, what is the distance from the center where the deflection is 1 cm?

Correct Answer: C

Q9.A rod AB of length 15 cm rests between two coordinate axes such that the endpoint A lies on the x-axis and the endpoint B lies on the y-axis. A point P(x, y) is taken on the rod such that AP = 6 cm. What is the length of PB?

Correct Answer: A

Q10.What is the standard form of the equation of a circle with a radius of 5 units?

Correct Answer: A

Q11.What is the length of the latus rectum of a hyperbola with the equation y2=4(5)xy^2 = 4(5)xy2=4(5)x?

Correct Answer: A

Q12.What is the definition of a circle?

Correct Answer: A

Q13.A circle is defined with a center at (3,4)(3, 4)(3,4) and a radius of 5 units. What is the equation of the circle?

Correct Answer: A

Q14.An ellipse has its major axis along the x-axis with vertices at (±6,0)(\pm 6, 0)(±6,0) and foci at (±4,0)(\pm 4, 0)(±4,0). What is the length of the minor axis?

Correct Answer: A

Q15.If a hyperbola has the equation x2a2−y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1a2x2​−b2y2​=1 and the length of the latus rectum is 10 units, what is the value of b2b^2b2 given a=5a = 5a=5?

Correct Answer: D

Q16.A rod AB of length 15 cm rests between two coordinate axes such that the endpoint A is on the x-axis and the endpoint B is on the y-axis. A point P(x, y) is on the rod such that AP = 6 cm. What is the locus of point P?

Correct Answer: B

Q17.A hyperbola has the equation x225−y216=1\frac{x^2}{25} - \frac{y^2}{16} = 125x2​−16y2​=1. What is the length of the transverse axis?

Correct Answer: B

Q18.If the equation of an ellipse is given by 9x2+16y2=1449x^2 + 16y^2 = 1449x2+16y2=144, what is the length of the semi-major axis?

Correct Answer: B

Q19.In a right triangle, if one angle is 30∘30^\circ30∘ and the hypotenuse is 10 units, what is the length of the side opposite to the 30∘30^\circ30∘ angle?

Correct Answer: A

Q20.What is the eccentricity of an ellipse with the equation x225+y29=1\frac{x^2}{25} + \frac{y^2}{9} = 125x2​+9y2​=1?

Correct Answer: B

Q21.An ellipse has its major axis along the x-axis and passes through the points (4,3) and (6,2). What is the equation of the ellipse?

Correct Answer: A

Q22.A rod of length 15 cm is placed between two coordinate axes such that one end lies on the x-axis and the other on the y-axis. If a point P divides the rod in the ratio 2:3, what are the coordinates of P?

Correct Answer: A

Q23.If the length of the latus rectum of a hyperbola is given by 2b2a\frac{2b^2}{a}a2b2​, what is the latus rectum for a hyperbola with a=3a = 3a=3 and b=4b = 4b=4?

Correct Answer: B

Q24.Given the ellipse equation x249+y236=1\frac{x^2}{49} + \frac{y^2}{36} = 149x2​+36y2​=1, what is the eccentricity of the ellipse?

Correct Answer: B

Q25.What is the equation of the ellipse with vertices at (0,±13)(0, \pm 13)(0,±13) and foci at (0,±5)(0, \pm 5)(0,±5)?

Correct Answer: B

Q26.What is the equation of a circle with a center at the origin and a radius of 5 units?

Correct Answer: A

Q27.In the context of conic sections, which statement is true about a parabola?

Correct Answer: A

Q28.A beam is deflected in the shape of a parabola with a deflection of 3 cm at the center. How far from the center is the deflection 1 cm?

Correct Answer: C

Q29.A rod of length 15 cm rests between the x-axis and y-axis. If a point P(x,y)P(x, y)P(x,y) on the rod is such that AP=6AP = 6AP=6 cm, where AAA is on the x-axis, what is the length of PBPBPB?

Correct Answer: D

Q30.If the length of the major axis of an ellipse is 26 and the foci are at (±5, 0), what is the value of the semi-major axis aaa?

Correct Answer: A

Q31.If a hyperbola has its foci at (±5,0)(\pm 5, 0)(±5,0) and the length of its latus rectum is 12, what is the equation of the hyperbola?