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Identifying Conic Sections from their Equation
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Identifying Conic Sections from their Equation

Last Updated : 12 Aug, 2024
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Conic section is defined as a locus of a point (P) in a plane which moves in such a way, that the ratio of its distances from a fixed point (called focus of conic, say S) and from a fixed line (called directrix of conic, say y-axis L) is constant (called the eccentricity of conic, denoted by 'e') i.e.

conic


PS / PM = constant (eccentricity e)

In other words, a Conic section are curves formed by the intersection of a plane with a right circular cone at different angles. The type of curve depends on the angle of intersection, resulting in different shapes such as circles, parabolas, ellipses, and hyperbolas.

Learn More, Conic sections

Types of Conic Sections

We can obtain the following curves from the intersections: 

  1. Circle
  2. Parabola
  3. Hyperbola
  4. Ellipse

There are certain characteristics that are common to each curve's equations. Let's study each of them and see how we can identify them from their equations. 

Circle

A circle is defined as the locus of a point which moves in a plane such that its distance from a fixed point in that plane is always constant.

Learn More: Circle in Math

The figure below represents a circle whose (fixed point) centre is given by O and the (constant distance) radius is the line joining the centre to any point on the circle. 

Its equation is given by,

(x - h)2 + (y - k)2 = r2

Where (h, k) is the centre of the circle, and the radius is given by "r".

General Equation of the Circle is given by ,

x2 + y2 -2hx -2ky + h2 + k2  = r2

Identifying Circles from their Equations

i) Equation of circle centred at x-axis:

x2 + y2 + 2hx + c = 0 , centre ( -h , 0 )

ii) Equation of circle centred at y-axis:

x2 + y2 + 2ky + c = 0 , centre ( 0 , -k )

iii) Equation of circle centred at origin:

x2 + y2 = r2 centre ( 0 , 0 )

iv) Equation of circle passing through at origin:

x2 + y2 + 2hx + 2ky = 0

v) Equation of circle in diametric form:

(x - x1)(x - x2) + (y - y1)(y - y2) = 0

vi) Equation of circle touching both the axes:

(x±r)2 + (y±r)2 = r2 centre (±r, ±r)

vii) Parametric form Equation of circle:

x = rcos𝞡 , y= rsin 𝞡 centre ( 0 , 0 )

For Example:

The equation given is, 4x2+ 4y2+ 7y= 9, The easiest way to identify when the given equation is the equation of a circle,

1. Both squares of x and y are present in the equation.

2. The coefficients of the squares of x and y are same (here, +4).

These two information tells that the provided curve's equation is a Circle.

Parabola

A parabola is the locus of a point which moves in a plane such that its distance from a fixed point (i.e. focus) is always equal to its distance from a fixed straight line (i.e. directrix).

Learn More, Parabola

Standard equations of Parabola

Equation of a standard parabola with x-axis as its axis, passing through the origin and focus at (a,0) is given by, 

y2 = 4ax

Identifying Parabolas from its Equations

i) Parabola towards right:

Equation of Parabola

y 2 = 4ax

Vertex

(0 , 0)

Focus

(a , 0)

Equation of directrix

x + a = 0

Parametric co-ordinates

(at2, 2at)

ii) Parabola towards left:

Equation of Parabola

y2 = -4ax

Vertex

(0,0)

Focus

(-a,0)

Equation of directrix

x - a = 0

Parametric co-ordinates

(-at2 , 2at)

iii) Parabola opening downwards:

Equation of Parabola

x2 = -4ay

Vertex

(0,0)

Focus

(0,-a)

Equation of directrix

y - a = 0

Parametric co-ordinates

(2at , -at2)

iv) Parabola opening upwards:

Equation of Parabola

x2 = 4ay

Vertex

(0,0)

Focus

(0,a)

Equation of Directrix

y + a = 0

Parametric co-ordinates

(2at , at2)

For Example:

Take a look at these equations,

x2= y+ 4, y2- 3x+ 9= 0

Both the equations mentioned above are a Parabola because,

1. Both equations have one of the variables squared, but not both. Therefore, the equations are parabola.

Ellipse 

An ellipse is the locus of a point which moves such that its distance from a fixed point (focus) is in a constant ratio to its distance from a fixed line (directrix). This ratio is called eccentricity e. (e<1 for ellipse)

The two points mentioned in the definition above are called foci of the ellipse.

Learn More, Ellipse

The standard equation of the ellipse that has the x-axis as the major axis and the y-axis as the minor axis is given by, 

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

where, c2 = a2 - b2

In case the ellipse has a major axis on the y-axis and so on. Then the equation is given by, 

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

Identifying Ellipses from their Equations

As similar to circles, ellipses also have x and y squares. But the difference is they will have different coefficients. Shape of Ellipse can be two types depending on which co-ordinate axis act as a major axis.

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

a>b

a<b

Centre

(0,0)

(0,0)

Vertices

(±a,0)

(0,±b)

Length of major axis

2a

2b

Length of minor axis

2b

2a

Foci

(±ae,0)

(0,±be)

Equation of directrices

x = ±a/e

y = ±b/e

For Example:

5x2+ 7y2- 9x- 6y = 0, 9y2+ 2y2+ 8x+ y= 0

Both the equations are the curves of ellipse because,

1. Both x and y variables are squared.

2. The coefficients of both squared variables are different in values (if the coefficients were equal, the curve would be a Circle)

Hyperbola

An Hyperbola is the locus of a point which moves such that its distance from a fixed point (focus) is in a constant ratio to its distance from a fixed line (directrix). This ratio is called eccentricity e. (e>1 for hyperbola)

Learn More, Hyperbola

The standard equation of Hyperbola is given by, 

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

Here also, c2 = a2 - b2

In the above case, the transverse axis is the x-axis, and the conjugate axis is the y-axis. If the axes are reversed, the equation will turn out to be, 

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

Identifying Hyperbolas from their Equations

As similar to circles, hyperbolas also have x and y squares. But the difference is they will have different coefficients and the signs will be opposite. 

Hyperbola

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

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

Centre

(0,0)

(0,0)

Length of transverse axis

2a

2b

Length of conjugate axis

2b

2a

Foci

(±ae,0)

(0, ±be)

Equation of directrices

x = ± a/e

x = ± b/e

For Example:

5x2- 2y2+ 7x= 8

The above equation is the equation of a Hyperbola since,

1. Both x and y with a degree 2 present.

2. one of the squared variables have different sign than other, in this case y2 coefficient is negative and x2 coefficient is positive.

Important Conic Sections Class 11 Resources:

In Class 11, studying conic sections is a key part of the mathematics curriculum, particularly within coordinate geometry. Conic sections—comprising circles, ellipses, parabolas, and hyperbolas—are important because they lay the groundwork for comprehending various physical phenomena. They have many practical applications in real life, such as in the orbits of planets and satellites, optics, and engineering designs.

  • Conic Sections Class 11 Notes
  • Conic Sections Class 11 NCERT Solutions

Let's see some sample problems on these concepts:

Sample Problems 

Question 1: Identify the curve from it's expanded equation. 

y2 -4y + 2 = 12x

Solution: 

y2 - 4y + 2 = 12x

This equation contains only y square. We have seen in the previous sections that the parabola equations have either x or y squared. So, this must be the equation of the parabola. 

Rearranging the given equation, 

y2 - 4y + 2 = 12x

⇒ y2 - 4y + 4 + 2 - 4 = 12x 

⇒(y - 2)2 - 2 =12x 

⇒(y - 2)2 = 12x + 2 

⇒(y - 2)^2= 12(x - \frac{1}{6})

Question 2: Identify the curve from its expanded equation: 

4x2 + 9y2 = 36

Solution: 

4x2 + 9y2 = 36

The given equation has both x and y squares present in it. Both have positive but different coefficients. So, it might be an equation of ellipse. 

4x^2 + 9y^2 = 36 \\ \frac{4x^2}{36} + \frac{9y^2}{36} = 1  \\ \frac{x^2}{9} + \frac{y^2}{4} =  1 \\ \frac{x^2}{3^2} + \frac{y^2}{2^2} = 1  

So, this is the equation with a = 3 and b = 2. 

Question 3: Identify the curve from its expanded equation: 

7x2 - 9y2 = 36

Solution: 

7x2 - 9y2 = 36

This equation also has squares of both x and y, but the signs are different. Based on the above-mentioned method, we can say that this is a hyperbola. 

Now we need to bring the expanded equation in standard form. 

7x^2 - 9y^2 = 36 \\ = \frac{7x^2}{36} -\frac{9y^2}{36} = 1 \\ = \frac{x^2}{\frac{36}{7}} - \frac{y^2}{\frac{36}{9}} = 1 \\ = \frac{x^2}{(\frac{6}{\sqrt{7}})^2} - \frac{y^2}{2^2} = 1 \\

Here, a = \frac{6}{\sqrt{7}}      and b = 2

Question 4: Given the expanded equation of a curve, identify it and bring it back to the standard form. 

x2 + y2  + 6y = 27

Solution: 

x2 + y2 + 6y = 27

Both x and y squares are present and have same sign and 1 as their coefficient. This is an equation of a circle. 

x2 + y2 + 6y = 27

⇒ x2 + y2 + 6y + 9 = 27 + 9 

⇒ x2 + (y + 3)2 = 36 

⇒ x2 + (y + 3)2 = 62

This is the equation of circle with centre at (0,-3) and radius 6.  

Question 5: Identify the curve and formulate its equation from the given expression. 

x2 + y2 + 4x + 6x = 12

Solution: 

Let's take the given equation,

x2 + y2 + 4x + 6x = 12

From the equations we have studied above, notice that in the given equation x and y both are squared and both have the same sign and same coefficients. As mentioned previously, equations of circle should have both x and y squares with same sign and coefficients. Thus, this is the equation of the circle. 

To find the centre and the radius of the circle, we need to rearrange it. 

x2 + y2 + 4x + 6x = 12

We need to make whole squares out of the x and y terms 

x2 + 4x + y2 + 6x = 12

⇒ x2 + 4x + 4 - 4 + y2 + 6x + 9 - 9 = 12

⇒ (x + 2)2 -4 + (y + 3)2 - 9 = 12 

⇒ (x + 2)2 + (y + 3)2 = 12 + 9 + 4 

⇒(x + 2)2 + (y + 3)2 = 25 

⇒(x + 2)2 + (y + 3)2 = 52

So, the centre is (-2,-3) and the radius is 5. 

Practice Problems on Identifying Conic Sections from their Equation

Question 1: Determine the equation for the ellipse that satisfies the given conditions: Centre at (0, 0), the major axis on the y-axis and passes through the points (3, 2) and (1, 6).

Question 2: Determine the equation of the hyperbola which satisfies the given conditions: Foci (0, ±13), the conjugate axis is of length 24.

Question 3: Find the equation of the circle which touches x-axis and whose centre is (1,2)

Question 4: Find the equations of the directrix & the axis of the parabola ⇒3x2=8y

Question 5: Show that the equation x2+y2−6x+4y−36=0 represents a circle, also find its centre & radius?

Question 6: Find the equation of an ellipse whose vertices are (0,±10) and e=4/5.

Question 7: Find the equation of the hyperbola with centre at the origin, length of the transverse axis 8 and one focus at (0,6)

Summary

Conic sections are curves derived from the intersection of a plane with a right circular cone at various angles, resulting in different shapes: circles, parabolas, ellipses, and hyperbolas. Identifying these conic sections from their equations involves recognizing the specific form of each type.


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