Mathematics CONIC SECTIONS -Introduction , Sections of a Cone , Degenerated conic sections ,Circle
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### Topics covered

• Introduction
• Sections of a Cone
• Circle, ellipse, parabola and hyperbola
• Degenerated conic sections
• Circle

### Introduction

\color{green} ✍️ In this Chapter, we shall study about some other curves, viz., circles, ellipses, parabolas and hyperbolas.

\color{green} ✍️ The names parabola and hyperbola are given by Apollonius.

\color{green} ✍️ These curves are in fact, known as conic sections or more commonly conics because they can be obtained as intersections of a plane with a double napped right circular cone.

\color{green} ✍️ These curves have a very wide range of applications in fields such as planetary motion, design of telescopes and antennas, reflectors in flashlights and automobile headlights, etc.

\color{green} ✍️ Now, in the subsequent sections we will see how the intersection of a plane with a double napped right circular cone results in different types of curves.

### Sections of a Cone

\color{green} ✍️ Let l be a fixed vertical line and m be another line intersecting it at a fixed point V and inclined to it at an angle α \
\ (Fig11.1).

• Suppose we rotate the line m around the line l in such a way that the angle α remains constant.

• Then the surface generated is a double-napped right circular hollow cone herein after referred as cone and extending indefinitely far in both directions (Fig11.2).

\color{green} ✍️ The point V is called the color{blue}(ul"vertex"); the line l is the axis of the cone.

\color{green} ✍️ The rotating line m is called a color{blue}(ul"generator of the cone.")

\color{green} ✍️ The vertex separates the cone into two parts called color{blue}(ul"nappes.")

• If we take the intersection of a plane with a cone, the section so obtained is called color{blue}(ul"a conic section.")

• Thus, conic sections are the curves obtained by intersecting a right circular cone by a plane.

• We obtain different kinds of conic sections depending on the position of the intersecting plane with respect to the cone and by the angle made by it with the vertical axis of the cone.

• Let β be the angle made by the intersecting plane with the vertical axis of the cone (Fig11.3).

• The intersection of the plane with the cone can take place either at the vertex of the cone or at any other part of the nappe either below or above the vertex.

### Circle, ellipse, parabola and hyperbola

When the plane cuts the nappe (other than the vertex) of the cone, we have the following situations:

(a) When β = 90^o, the section is a color{blue}(ul"circle") (Fig11.4).

(b) When α < β < 90^o, the section is an color{blue}(ul"ellipse") (Fig11.5).

(c) When β = α , the section is a color{blue}(ul"parabola") (Fig11.6).

(In each of the above three situations, the plane cuts entirely across one nappe of the cone).

(d) When 0 ≤ β < α; the plane cuts through both the nappes and the curves of intersection is a color{blue}(ul"hyperbola") (Fig11.7).

### Degenerated conic sections

When the plane cuts at the vertex of the cone, we have the following different cases :

(a) When α < β ≤ 90^o, then the section is a color{blue}(ul"point") (Fig11.8).

(b) When β = α, the plane contains a generator of the cone and the section is a color{blue}(ul"straight line") (Fig11.9).
It is the degenerated case of a parabola.

(c) When 0 ≤ β < α, the section is a pair of intersecting straight lines (Fig11.10).

### Circle

color{blue}(★ ul"Circle")

\color{green} ✍️ A color{blue}(ul"circle") is the set of all points in a plane that are equidistant from a fixed point in the plane.

\color{green} ✍️ The fixed point is called the color{blue}(ul"centre") of the circle and the distance from the centre to a point on the circle is called the color{blue}(ul"radius") of the circle (Fig 11.11).

color{green}(ul"The equation of the circle is simplest if the centre of the circle is at the origin.")

However, we derive below the equation of the circle with a given centre and radius (Fig 11.12).

Given C\ \ (h, k) be the centre and r the radius of circle. Let P(x, y) be any point on the circle (Fig11.12).

Then, color{blue}("by the definition"), | CP | = r . By the distance formula, we have

i.e sqrt((x-h)^2 + (y-k)^2=r

color {green}{ul[(x-h)^2 + (y-k)^2 = r^2]}

This is the required equation of the circle with centre at (h,k) and radius r .
Q 3175591466

Find an equation of the circle with centre at (0,0) and radius r.

Solution:

Here h = k = 0. Therefore, the equation of the circle is x^2 + y^2 = r^2.
Q 3105591468

Find the equation of the circle with centre (–3, 2) and radius 4.

Solution:

Here h = –3, k = 2 and r = 4. Therefore, the equation of the required circle is (x + 3)^2 + (y –2)^2 = 16
Q 3115691560

Find the centre and the radius of the circle x^2 + y^2 + 8x + 10y – 8 = 0

Solution:

The given equation is
(x^2 + 8x) + (y^2 + 10y) = 8

Now, completing the squares within the parenthesis, we get
(x^2 + 8x + 16) + (y^2 + 10y + 25) = 8 + 16 + 25
i.e. (x + 4)^2 + (y + 5)2 = 49
i.e. {x – (– 4)}^2 + {y – (–5)}^2 = 72
Therefore, the given circle has centre at (– 4, –5) and radius 7.
Q 3135691562

Find the equation of the circle which passes through the points (2, – 2), and (3,4) and whose centre lies on the line x + y = 2.

Solution:

Let the equation of the circle be (x – h)^2 + (y – k)^2 = r^2.

Since the circle passes through (2, – 2) and (3,4), we have

(2 – h)^2 + (–2 – k)^2 = r^2 ... (1)
and (3 – h)^2 + (4 – k)^2 = r^2 ... (2)
Also since the centre lies on the line x + y = 2, we have
h + k = 2 ... (3)
Solving the equations (1), (2) and (3), we get
h = 0.7, k = 1.3 and r^2 = 12.58
Hence, the equation of the required circle is
(x – 0.7)2 + (y – 1.3)^2 = 12.58.