Class 11 Mathematics Chapter 10 Conic Sections

Key Concept: Eccentricity Calculation

b) 1.28

[Solution Description]

The eccentricity $e$ is calculated using the formula $e = c/a$, where $c^2 = a^2 + b^2$. Here, $a^2 = 25$ and $b^2 = 16$.

Calculate $c$:

$$c = \sqrt{25 + 16}$$

$$= \sqrt{41}$$

Given that $a = 5$, the eccentricity becomes:

$$e = \frac{\sqrt{41}}{5}$$

Thus, $e = \frac{\sqrt{41}}{5}$ which approximately equals $1.28$.

Key Concept: Basic Definition

d) A curve obtained by intersecting a plane with a double-napped right circular cone

[Solution Description]

A conic section is a curve obtained by intersecting a plane with a double-napped right circular cone. The correct choice corresponds to this definition.

Key Concept: Section Type

d) Circle

[Solution Description] A plane that cuts parallel to the base of a right circular cone forms a circle. This happens because the angle $\beta$ between the plane and the vertical axis is equal to 90°.

Key Concept: Transverse Axis Identification

b) 4

[Solution Description]

The standard equation of the hyperbola is $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$, indicating that the transverse axis is along the x-axis.

Here, $a^2 = 4$, thus $a = 2$.

The length of the transverse axis is $2a = 2 \times 2 = 4$.

Key Concept: Basic Definition

c) Parabola

[Solution Description] The design of automobile headlights often uses a parabolic reflector to focus and direct light effectively. A parabola has reflective properties that allow it to collect or distribute energy such as light.

Key Concept: Historical Context

c) Apollonius

[Solution Description] Apollonius is known for his work called ‘The Conic’, which has been a fundamental piece in understanding and developing the theory of conic sections.

Key Concept: Conceptual Understanding

b) A pair of intersecting lines

[Solution Description] The condition given states $0 \leq \beta < \alpha$. According to the syllabus, this results in a pair of intersecting straight lines, which is called a degenerated hyperbola.

Key Concept: Locus Problems

c) $x^2 + y^2 – 4y + 4 = 16y^2$

[Solution Description]

Let the point be $(x, y)$. Its distance from $(0, 2)$ is $\sqrt{x^2 + (y – 2)^2}$ and its distance from the line $y = 0$ is $|y|$. According to the problem condition:

$$\sqrt{x^2 + (y – 2)^2} = 4|y|$$

Squaring both sides:

$$x^2 + (y – 2)^2 = 16y^2$$

Simplifying:

$$x^2 + y^2 – 4y + 4 = 16y^2$$

Rearrange terms:

$$x^2 + y^2 – 4y + 4 = 16y^2 \implies x^2 + y^2 – 4y + 4 – 16y^2 = 0$$

Therefore, the equation of the locus is $x^2 + y^2 – 4y + 4 = 16y^2$.

Key Concept: Straight Line Condition

b) $\beta = \alpha$

[Solution Description] The section of a cone turns into a straight line, which is the degenerated case of a parabola, if the angle of intersection satisfies $\beta = \alpha$. This means the cutting plane is parallel to the slant height of the cone and only touches one side along its entire length.

Key Concept: Real-World Application

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

[Solution Description]

The standard form of the equation of an ellipse aligned along the x-axis is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ is the semi-major axis. The distance of the foci from the center $c = \pm 3$ and $a = 5$.

Using the relationship $c^2 = a^2 – b^2$, solve for $b^2$:

$$c = 3, \quad a = 5$$

$$c^2 = a^2 – b^2$$

$$b^2 = a^2 – c^2$$

$$= 5^2 – 3^2$$

$$= 25 – 9$$

$$= 16$$

Therefore, the equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{16} = 1$.

Key Concept: Standard Equation

b) $\frac{y^2}{25} – \frac{x^2}{b^2} = 1$

[Solution Description] For a hyperbola centered at the origin with foci on the y-axis, its standard form is $\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$. If the length of the transverse axis is 10, then $2a = 10$, which means $a = 5$. Therefore, the standard equation is:

$$\frac{y^2}{25} – \frac{x^2}{b^2} = 1.$$

Since $b^2$ is not specified in the problem, it remains as is for the standard form.

Key Concept: Eccentricity Calculation

a) 5/4

[Solution Description] The standard form of a hyperbola is $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$, where $e = \frac{c}{a}$. For the given equation, $a^2 = 16$, so $a = 4$. Also, $b^2 = 9$, so $b = 3$. We calculate $c$ using $c^2 = a^2 + b^2$. Thus, $c^2 = 16 + 9 = 25$, giving $c = \sqrt{25} = 5$. Now compute the eccentricity as $e = \frac{c}{a} = \frac{5}{4}$.

Key Concept: Diameter Calculation

b) 10

[Solution Description]

The equation $x^2 + y^2 = 25$ indicates $r^2 = 25$. Solving for $r$, we get:

$$r = \sqrt{25} = 5.$$

The diameter of the circle is twice the radius, so:

$$d = 2 \times 5 = 10.$$

Hence, the diameter is 10.

Key Concept: Eccentricity Calculation

b) $\frac{\sqrt{11}}{6}$

[Solution Description]

The semi-major axis $a$ is half of the major axis length: $a = 12/2 = 6$. The semi-minor axis $b$ is half of the minor axis length: $b = 10/2 = 5$. The eccentricity is calculated using:

$$e = \frac{c}{a}$$

where $c^2 = a^2 – b^2$. Therefore,

$$c^2 = 6^2 – 5^2 = 36 – 25 = 11$$

$$c = \sqrt{11}$$

Hence, the eccentricity is:

$$e = \frac{\sqrt{11}}{6}$$

Key Concept: Axis of Symmetry

a) x-axis

[Solution Description] The equation $y^2 = 4ax$ represents a parabola with its axis along the x-axis. The symmetry of such a parabola is also along the x-axis because the square term is associated with the y variable.

Key Concept: Basic Definition

d) A set of points where the sum of distances from two fixed points is constant.

[Solution Description] An ellipse is defined as the set of all points such that the sum of the distances from two fixed points (called foci) is constant.

Key Concept: Basic Equation

b) $$\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$$

[Solution Description] The standard equation for a hyperbola with its transverse axis along the y-axis is $$\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$$. This is derived by positioning the transverse axis vertically, which changes the orientation.

Key Concept: Conditional Reasoning

c) Point

[Solution Description]

When $\alpha < \beta \leq 90^\circ$, the resulting section is a point. Here, $\alpha = 45^\circ$ and $\beta = 50^\circ$, which satisfies the condition for forming a point. Therefore, the answer is: Point.

Key Concept: Ellipse Latus Rectum

b) $\frac{16}{3}$

[Solution Description] For an ellipse of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the length of the latus rectum is $\frac{2b^2}{a}$. Here $a^2 = 36$, so $a = 6$, and $b^2 = 16$, thus $b = 4$. The length of the latus rectum is:

$$\frac{2(16)}{6} = \frac{32}{6} = \frac{16}{3}.$$

Key Concept: Identify Center

b) (5, -3)

[Solution Description] The equation of the circle is given as $(x – 5)^2 + (y + 3)^2 = 25$. Here, comparing with the general form $(x – h)^2 + (y – k)^2 = r^2$, we find $h = 5$ and $k = -3$. Therefore, the center of the circle is $(5, -3)$.

Key Concept: Latus Rectum Length

d) 16

[Solution Description] For the parabola $y^2 = 4ax$, the length of the latus rectum is $4a$. Here, $4a = 16$, so $a = 4$. Therefore, the length of the latus rectum is $4 \times 4 = 16$.

Key Concept: Coordinate Calculation

a) (8, 4) and (-8, 4)

[Solution Description]

The given equation is $x^2 = 16y$. Comparing with the standard form $x^2 = 4ay$, we have $4a = 16$, which gives $a = 4$.

The length of the latus rectum for a parabola is $4a = 16$.

The vertex is at $(0,0)$ and focus at $\left(0,\frac{a}{1}\right)= (0,4)$. Since the parabola opens upwards, the endpoints of the latus rectum are at equal distance from the focus along the $x$-axis.

So the endpoints are $(8,4)$ and $(-8,4)$.

Key Concept: Symmetry Understanding

a) Transverse axis

[Solution Description] The latus rectum of a hyperbola is symmetric with respect to its transverse axis.

Key Concept: Basic Definition

c) 5/3

[Solution Description] To find the eccentricity $e$ of a hyperbola, we use the formula $e = c/a$. Given $c = 10$ and $a = 6$, we calculate:

$$e = \frac{c}{a} = \frac{10}{6} = \frac{5}{3}.$$

So, the eccentricity is $\frac{5}{3}$.

Key Concept: Equation-Derivation

a) $(x – 3)^2 + (y – 3)^2 = 8$

[Solution Description]

To derive the equation of the circle, use the formula $(x – h)^2 + (y – k)^2 = r^2$, where $(h, k)$ is the center.

Given center is $(3, 3)$. Use point $(1, 5)$ to find the radius:

$$(1 – 3)^2 + (5 – 3)^2 = r^2$$

$$(-2)^2 + 2^2 = r^2$$

$$4 + 4 = r^2$$

Therefore, $r^2 = 8$. The equation becomes:

$$(x – 3)^2 + (y – 3)^2 = 8$$

Key Concept: Basic Identification

d) Hyperbola

[Solution Description]

For an angle $\beta = 45^\circ$, we know that it is less than $\alpha$ ($\alpha < \beta \leq 90^\circ$ for ellipse). Therefore, since $0 \leq \beta < \alpha$, this falls under the category of a hyperbola.

Key Concept: Types of Conic Sections

a) Circle

[Solution Description] When the angle $\beta = 90^\circ$, the plane intersection is perpendicular to the vertical axis, and the shape formed is a circle.

Key Concept: Basic Definition

b) A set of points equidistant from a fixed point

[Solution Description] A circle can be defined as the set of all points in a plane that are equidistant from a fixed point, which is known as the center.

Key Concept: Types of Conic Sections

b) Circle

[Solution Description] A conic section formed when the angle $\beta$ is $90^\circ$ with the vertical axis is a circle.

Key Concept: Latus Rectum Length

d) 16

[Solution Description]

The given equation is $y^2 = 4ax$, where $4a = 16$. Solving gives $a = 4$. The length of the latus rectum of a parabola in this form is $4a$. Therefore, the length is $4 \times 4 = 16$ units.

Key Concept: Basic Definitions

b) Ellipse

[Solution Description] The paths of planets around the sun are elliptical in shape due to gravitational forces and their motion, which fits the properties of an ellipse.

Key Concept: Basic Definition

b) 1.25

[Solution Description]

The eccentricity $e$ of a hyperbola is calculated using the formula $e = \frac{c}{a}$.

Given values:

– $c = 10$

– $a = 8$

Substitute these into the formula:

$$e = \frac{10}{8} = \frac{5}{4} = 1.25$$

Therefore, the eccentricity of the hyperbola is $1.25$.

Key Concept: Equation Identification

a) $(x – h)^2 + (y – k)^2 = r^2$

[Solution Description] The standard form of the equation of a circle with center $(h, k)$ and radius $r$ is given by $(x – h)^2 + (y – k)^2 = r^2$.

Key Concept: Parabola Definition

c) The set of all points equidistant from a fixed line and a fixed point not on the line

[Solution Description] A parabola is defined as the set of all points in a plane that are equidistant from a fixed line, called the directrix, and a fixed point not on the line, known as the focus.

Key Concept: Eccentricity Calculation

b) 0.6

[Solution Description]

The formula for eccentricity $e$ is given by $e = \frac{c}{a}$, where $c^2 = a^2 – b^2$. Here, $a = 10$ and $b = 8$.

Step 1: Calculate $c$.

$$c^2 = a^2 – b^2 = 10^2 – 8^2 = 100 – 64 = 36$$

$$c = \sqrt{36} = 6$$

Step 2: Calculate $e$.

$$e = \frac{c}{a} = \frac{6}{10} = 0.6$$

Thus, the eccentricity of the ellipse is 0.6.

Key Concept: Equation Derivation

a) $(x – 3)^2 + (y + 4)^2 = 25$

[Solution Description]

To find the equation of the circle, we use the standard formula for a circle $\left(x – h\right)^2 + \left(y – k\right)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.

Given: Centre $(h, k) = (3, -4)$ and radius $r = 5$

Substitute these values into the formula:

$$\left(x – 3\right)^2 + \left(y + 4\right)^2 = 5^2$$

Simplifying further,

$$\left(x – 3\right)^2 + \left(y + 4\right)^2 = 25$$

Therefore, the equation is $\left(x – 3\right)^2 + \left(y + 4\right)^2 = 25$.

Key Concept: Complex Angle Scenarios, Real-world Application

c) Ellipse

[Solution Description] Since the horizontal plane intersects at $\beta = 45^\circ$, which is less than $90^\circ$ but greater than $\alpha$, it forms an ellipse according to the conic section definitions.

Key Concept: Basic Definition

d) Hyperbola

[Solution Description] A hyperbola is formed when a plane intersects both nappes of a double cone.

Key Concept: Parabola Calculation

b) 20

[Solution Description] The standard form of a vertical parabola is $x^2 = 4ay$. Here, comparing $x^2 = 20y$, we find $4a = 20$, hence $a = 5$. Thus, the length of the latus rectum is $4a = 4 \times 5 = 20$.

Key Concept: Application of Formula

b) 0.8

[Solution Description]

The eccentricity $e$ of an ellipse is given by $e = \frac{c}{a}$, where $c$ is the distance from the center to one of the foci and $a$ is the semi-major axis.

Given that $c = 4$ and $a = 5$, we can plug these values into the formula:

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

Therefore, the eccentricity is 0.8.

Key Concept: Equation Formation

a) $(x – 5)^2 + (y + 3)^2 = 36$

[Solution Description]

The equation of a circle with center $(h, k)$ and radius $r$ is given by:

$$(x – h)^2 + (y – k)^2 = r^2$$

Here, $h = 5$, $k = -3$, and $r = 6$. Substituting these values into the equation gives:

$$(x – 5)^2 + (y + 3)^2 = 36$$

Key Concept: Axis Identification

b) y-axis

[Solution Description] The given hyperbola is in the form $\dfrac{y^2}{a^2} – \dfrac{x^2}{b^2} = 1$, indicating that the transverse axis is along the y-axis since the positive term involves $y^2$.

Key Concept: Circle Properties

c) Center: (-4, 5), Radius: 7

[Solution Description] 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. Comparing this to the given equation $(x + 4)^2 + (y – 5)^2 = 49$, we identify that:

– $h = -4$ (as it is $(x – (-4))$), so the x-coordinate of the center is $-4$.

– $k = 5$, so the y-coordinate of the center is $5$.

– $r^2 = 49$, hence $r = \sqrt{49} = 7$.

Therefore, the center is at $(-4, 5)$ and the radius is $7$

Key Concept: Major and Minor Axes

b) The longest diameter of the ellipse.

[Solution Description] The major axis of an ellipse is the longest diameter, which passes through the foci and the center of the ellipse.

Key Concept: Circle Definition

b) The set of all points in a plane equidistant from a fixed point

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

Key Concept: Equation Derivation

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

[Solution Description] The standard equation of an ellipse with foci on the x-axis is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ is the semi-major axis and $b$ is the semi-minor axis. Here, $a = 7$ and $b = 5$. Substituting these values into the equation gives us:

$$\frac{x^2}{7^2} + \frac{y^2}{5^2} = 1$$

Simplifying this, we get:

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

Key Concept: Identify Components

c) Vertex

[Solution Description] The vertex of a parabola is the midpoint between the focus and the directrix.

Key Concept: Equation Identification

b) $y^2 = 20x$

[Solution Description] The standard form for a parabola opening to the right with vertex at the origin is $y^2 = 4ax$. Given the focus at (5,0) and directrix x = -5, the value of $a$ (distance from vertex to focus) is 5. Thus, the equation becomes:

$$y^2 = 4 \times 5 \times x = 20x.$$

This means the equation of the parabola is $y^2 = 20x$.