# 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.

### What is Hyperbola?

Hyperbolas are similar to mirrored parabolas in appearance. The branches are the two parts of the tree. Hyperbola is generated when the plane crosses the halves of a right circular cone, the angle of which is parallel to the cone’s axis.

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.

**Properties of Hyperbola**

- In a hyperbola, the difference between the focal distances is constant, which is the same as the transverse axis ||PS – PS’|| = 2a.
- If e
_{1}and e_{2}are the eccentricities of hyperbola, the relation e_{1}^{-2}+ e_{2}^{-2}= 1 applies. - If the lengths of the transverse and conjugate axes are equal, the hyperbola is said to be rectangular or equilateral.
- The eccentricity of a rectangular hyperbola is √2, which is the same as the length of the axes’ latus rectum.
- If the point (x
_{1}, y_{1}) within, on, or outside of the hyperbola, the value of x_{1}^{2}/a^{2}– y_{1}2/ b^{2}= 1 is positive, zero, or negative. - Two lines intersect the center of the hyperbola. The tangents to the center are the hyperbola’s asymptotes.
- 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. 2b
^{2}/a is the answer.

### Equation of Hyperbola

The General Equation of the hyperbola is:

(x−x_{0})^{2}/a^{2 }− (y−y_{0})^{2}/b^{2 }= 1where, a is the semi-major axis and b is the semi-minor axis, x

_{0}, and y_{0}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 = √(c
^{2}–a^{2})

**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 Directrixor

e = c/a

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

e =√(1+b^{2}/a^{2})

**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 = y_{0}

**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 = x_{0}

**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 = y_{0}+ (b/a)x – (b/a)x_{0}

y = y_{0}– (b/a)x + (b/a)x_{0}

**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 = ±a^{2 }/√(a^{2}+ b^{2})

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

[a, y_{0 }] ; [-a, y_{0}]

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

(x_{0}+ √(a^{2}+ b^{2}), y_{0})

(x_{0}– √(a^{2}+ b^{2}), y_{0})

**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.

(xand^{2}/ a^{2}) – (y^{2}/b^{2}) = 1(−x^{2}/ a^{2}) + (y^{2}/ b^{2}) = 1are conjugate hyperbolas together each other. Therefore:

(y^{2}/ b^{2}) – (x^{2}/ a^{2}) = 1

a^{2}= b^{2}(e^{2}− 1)

e = √( 1+ a^{2}/b^{2})

### Sample Questions

**Question 1: Find the eccentricity of hyperbola having the equation x ^{2}/36 – y^{2}/49 = 1.**

**Answer: **

Given equation is x

^{2}/36 – y^{2}/49 = 1.Comparing with equation of the hyperbola x

^{2}/a^{2}– y^{2}/b^{2}= 1Where a

^{2 }= 36 and b^{2}= 49As we know formula for eccentricity,

e = √( 1+ b

^{2}/a^{2})= √( 1 + 49/ 36 )

= √( 36+49/36 )

= √(85/36)

= √85/√36

= 9.21/6

= 1.53

Therefore we got eccentricity for x

^{2}/36 – y^{2}/49 = 1 is1.53.

**Question 2: Find the eccentricity of hyperbola having the equation x ^{2}/27 – y^{2}/25 = 1.**

**Answer:**

Given equation is x

^{2}/27 – y^{2}/25 = 1Comparing with equation of the hyperbola x

^{2}/a^{2}– y^{2}/b^{2}= 1Where a

^{2}= 27 and b^{2}= 25As we know formula for eccentricity,

e = √( 1+ b

^{2}/a^{2})= √( 1 + 25/ 27 )

= √( 27+25/27 )

= √(52/27)

= √52/√27

= 7.21/5.19

= 1.38

Therefore we got eccentricity for x

^{2}/36 – y^{2}/49 = 1 is1.38.

**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+ b

^{2}/a^{2})then

1.3 = √(1+b

^{2}/20^{2})13/10 = √( 400 + b

^{2}/ 400 )(13/10)

^{2}= ( 400 +b^{2}/400 )169/100 = ( 400 +b

^{2}/400 )b

^{2}= 276Comparing with equation of the hyperbola x

^{2}/a^{2}– y^{2}/b^{2}= 1then,

x^{2}/400 – y^{2}/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 = ±a^{2}/√(a^{2}+ b^{2})

**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+ a^{2}/b^{2})

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