Hyperbola Formula: The set of all points in a plane is called a hyperbola. The distance between these two fixed points in the plane will remain constant. The distance to the distant location minus the distance to the nearest point is the difference. The foci will be the two fixed points, and the center of the hyperbola will be the mid-point of the line segment connecting the foci. Hyperbola is a fascinating topic in geometrical mathematics.

This article explores the hyperbola formulas, along with their equations, and solved examples on it.

What is Hyperbola?

A hyperbola is a set of points where the distance between each focus is always larger than one. To put it another way, the locus of a point moving in a plane when the distance between a fixed point (focus) and a fixed line (directrix) is a constant greater than 1.

A hyperbola is made up of two foci and two vertices. The foci of the hyperbola are placed away from the center and vertices. The line that goes through the foci is known as the transverse axis. The conjugate axis is perpendicular to the transverse axis and goes through the center. The vertices are the positions where the hyperbola crosses the transverse axis.

Read in Detail: Hyperbola

Properties of Hyperbola

1. In a hyperbola, the difference between the focal distances is constant, which is the same as the transverse axis ||PS – PS’|| = 2a.
2. If e₁ and e₂ are the eccentricities of hyperbola, the relation e₁^-2 + e₂^-2 = 1 applies.
3. If the lengths of the transverse and conjugate axes are equal, the hyperbola is said to be rectangular or equilateral.
4. The eccentricity of a rectangular hyperbola is √2, which is the same as the length of the axes’ latus rectum.
5. If the point (x₁, y₁) within, on, or outside of the hyperbola, the value of x₁²/a² – y₁2/ b² = 1 is positive, zero, or negative.
6. Two lines intersect the center of the hyperbola. The tangents to the center are the hyperbola’s asymptotes.
7. The latus rectum of a hyperbola is a line that runs perpendicular to the transverse axis and crosses through either of the conjugate axis’ parallel foci. 2b²/a is the answer.

Equation of Hyperbola

The General Equation of the hyperbola is:

(x−x₀)²/a² − (y−y₀)²/b² = 1

where, a is the semi-major axis and b is the semi-minor axis, x₀, and y₀ are the center points, respectively.

• The distance between the two foci would always be 2c.
• The distance between two vertices would always be 2a. It is also can be the length of the transverse axis.
• The conjugate axis will be 2b in length, here b = √(c²–a²)

Hyperbola Formulas

Hyperbolic Eccentricity Formula: A hyperbola’s eccentricity is always greater than 1, i.e. e > 1. The ratio of the distance of the point on the hyperbole from the focus to its distance from the directrix is the eccentricity of a hyperbola.

Eccentricity = Distance from Focus/Distance from Directrix 

or

e = c/a

We get the following value of eccentricity by substituting the value of c.

e =√(1+b²/a²)

Equation of Major axis: The Major Axis is the line that runs through the center, hyperbola’s focus, and vertices. 2a is considered the length of the major axis. The formula is as follows: 

y = y₀

Equation of Minor axis: The Minor Axis is a line that runs orthogonal to the major axis and travels through the middle of the hyperbola. 2b is the length of the minor axis. The following is the equation:

x = x₀

Asymptotes: The Asymptotes are two bisecting lines that pass through the center of the hyperbola but do not touch the curve. The following is the equation:

y = y₀ + (b/a)x – (b/a)x₀

y = y₀ – (b/a)x + (b/a)x₀

Directrix of a hyperbola: A hyperbola’s directrix is a straight line that is utilized to generate a curve. It is also known as the line away from which the hyperbola curves. The symmetry axis is perpendicular to this line. The directrix equation is:

x = ±a² /√(a² + b² )

Vertex: The vertex is the point on a stretched branch that is closest to the center. These are the vertex points.

[a, y₀ ] ; [-a, y₀ ]

Focus (foci): Focus ( foci) are the fixed locations on a hyperbola where the difference between the distances is always constant.

(x₀ + √(a² + b²), y₀)

(x₀ – √(a² + b²), y₀)

Conjugate Hyperbola: Two hyperbolas whose transverse and conjugate axes are the conjugate and transverse axes of the other are referred to as conjugate hyperbolas of each other.

(x² / a²) – (y² /b²) = 1 and  (−x² / a²) + (y² / b²) = 1 

are conjugate hyperbolas together each other. Therefore:

(y² / b²) – (x² / a²) = 1

a² = b² (e² − 1)

e = √( 1+ a²/b²)

Applications of Hyperbola

Hyperbolas have numerous applications in various fields such as astronomy, physics, and engineering. Some key applications include:

• Orbit Paths: In astronomy, the paths of comets around the sun often form hyperbolic trajectories.
• Navigation Systems: Hyperbolic navigation systems use the properties of hyperbolas for accurate positioning.
• Acoustics: Hyperbolic shapes are used in the design of certain types of acoustic mirrors and reflectors to focus sound waves.

Learn More:

• Parabola
• Ellipse
• Conic section

Sample Questions on Hyperbola Formula

Question 1: Find the eccentricity of hyperbola having the equation x²/36 – y²/49 = 1.

Answer: 

Given equation is x²/36 – y²/49 = 1.

Comparing with equation of the hyperbola x²/a² – y²/b² = 1

Where a² = 36 and b² = 49

As we know formula for eccentricity,

e = √( 1+ b²/a² )

= √( 1 + 49/ 36 )

= √( 36+49/36 )

= √(85/36)

= √85/√36

= 9.21/6

= 1.53

Therefore we got eccentricity for x²/36 – y²/49 = 1 is 1.53.

Question 2: Find the eccentricity of hyperbola having the equation x²/27 – y²/25 = 1.

Answer:

Given equation is  x²/27 – y²/25 = 1

Comparing with equation of the hyperbola x²/a² – y²/b² = 1

Where a² = 27 and b² = 25

As we know formula for eccentricity,

e = √( 1+ b²/a² )

= √( 1 + 25/ 27 )

= √( 27+25/27 )

= √(52/27)

= √52/√27

= 7.21/5.19

= 1.38

Therefore we got eccentricity for x²/36 – y²/49 = 1 is 1.38.

Question 3: A hyperbola has an eccentricity of 1.3 and the value of a is 20. Find the hyperbola’s equation.

Answer:

Given eccentricity is 1.38 and value for a is  20

As we know the formula for eccentricity,

e = √( 1+ b²/a² )

then

1.3 =  √(1+b²/20²)

13/10 =  √( 400 + b²/ 400 )

(13/10)² = ( 400 +b²/400 )

169/100 = ( 400 +b²/400 )

b²= 276

Comparing with equation of the hyperbola x²/a² – y²/b² = 1

then,

x²/400 – y²/276 = 1

Question 4: State what is a Hyperbola?

Answer:

The locus of a point moving in a plane where the ratio of its distance from a fixed point to that from a fixed-line is a constant greater than one is called a hyperbola.

Question 5: What is the directrix of hyperbola and its formula?

Answer: 

A hyperbola’s directrix is a straight line that is utilized to generate a curve. It is also known as the line away from which the hyperbola curves. The symmetry axis is perpendicular to this line. The directrix equation is:

x = ±a² /√(a² + b²)

Question 6: What is the formula for conjugate hyperbola?

Answer:

Two hyperbolas whose transverse and conjugate axes are the conjugate and transverse axes of the other are referred to as conjugate hyperbolas of each other.

e = √( 1+ a²/b² )

Practice Problems on Hyperbola Formula

Problem 1: Find the length of the transverse and conjugate axes for the hyperbola with the equation x²/16 – y²/9 =1.

Problem 2: Given the hyperbola (x-2)²/25 – (y+3)²/16 = 1, find the coordinates of the center, vertices, and foci.

Problem 3: Determine the equation of the asymptotes for the hyperbola x²/36 – y²/16 =1.

Problem 4: For the hyperbola x²/49 – y²/64 = 1, calculate the coordinates of the vertices and the equations of the directrices.

Problem 5: A hyperbola has its center at the origin, a transverse axis of length 10, and an eccentricity of 2. Find its standard equation.

Problem 6: Find the latus rectum of the hyperbola given by the equation (x-1)²/4 – (y-2)²/9 = 1.

Problem 7: Verify if the point (5, 3) lies on the hyperbola given by x²/25 – y²/9 =1.

Problem 8: Determine the distance between the foci of the hyperbola x²/81 – y²/64 =1.

Problem 9: Find the coordinates of the center, vertices, and foci for the hyperbola given by (x-3)²/49 – (y+4)²/25 = 1.

Problem 10: For the hyperbola with the equation x²/81 – y²/16=1, determine the equations of the asymptotes and the length of the latus rectum.

Summary

A hyperbola is a set of points where the absolute difference of distances from any point on the hyperbola to two fixed points (foci) is constant. The hyperbola has key components such as the center, transverse axis, conjugate axis, vertices, foci, and asymptotes. The standard equation for a hyperbola can be expressed in two forms, based on the orientation of the transverse axis. The properties of hyperbolas include eccentricity, which is always greater than 1, and the relationship between the lengths of the axes and the distances between key points like the foci and vertices. Hyperbolas are significant in various fields, including astronomy, physics, and engineering, for their unique geometric properties and applications.

FAQs on Hyperbola Formula

What is a hyperbola in simple terms?

A hyperbola is a type of conic section that forms when a plane intersects both nappes of a double cone. It consists of two open curves that are mirror images of each other.

How is a hyperbola different from an ellipse?

A hyperbola and an ellipse are both conic sections, but they differ in their geometric properties. An ellipse is the set of points where the sum of the distances from any point on the ellipse to two fixed points (foci) is constant. In contrast, a hyperbola is the set of points where the absolute difference of the distances from any point on the hyperbola to two fixed points (foci) is constant.

What is the eccentricity of a hyperbola?

The eccentricity (e) of a hyperbola is a measure of how “stretched” the hyperbola is. It is always greater than 1 and is given by the formula e = √(1+b²/a²) , where a and b are the semi-major and semi-minor axes, respectively.

How do you find the foci of a hyperbola?

The foci of a hyperbola are found using the distance c from the center, where c =√(a²+ b²). For a hyperbola centered at (x₀ , y₀), the coordinates of the foci are (x₀±c , y₀) for a horizontally oriented hyperbola and (x₀ , y₀±c) for a vertically oriented hyperbola.

What are the asymptotes of a hyperbola?

The asymptotes of a hyperbola are straight lines that pass through the center and define the directions in which the hyperbola opens. For a hyperbola (x² / a²) – (y² /b²) = 1 ,the equations of the asymptotes are y= ±(b/a)x. For a hyperbola (x² / a²) – (y² /b²) = -1, the equations of the asymptotes are y= ±(a/b)x.

Can a hyperbola be a circle?

No, a hyperbola cannot be a circle. A hyperbola has two branches that extend infinitely, while a circle is a closed curve with all points equidistant from the center.

How is the length of the latus rectum calculated?

The length of the latus rectum of a hyperbola is given by the formula 2(b²/a²), where a and b are the lengths of the semi-major and semi-minor axes, respectively.

What is the conjugate axis of a hyperbola?

The conjugate axis of a hyperbola is the axis perpendicular to the transverse axis and passes through the center. It measures the distance between the points where the hyperbola intersects a line perpendicular to the transverse axis at the center.

How do you determine if a point lies on a hyperbola?

To determine if a point (x₁ , y₁) lies on a hyperbola given by the equation x²/a² – y²/b² = 1, substitute x₁ and y₁ into the equation. If the equation holds true, then the point lies on the hyperbola.

What is the standard equation of a hyperbola centered at the origin?

The standard equation of a hyperbola centered at the origin is x²/a² – y²/b² = 1 for a horizontal hyperbola, and y²/b² – x²/a² = 1for a vertical hyperbola.