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# Equation of a Hyperbola

• Last Updated : 03 Jul, 2022

In mathematics, a hyperbola is a significant conic section that is formed when a plane surface intersects a double cone, but certainly not at the center. As a result of the intersection of a double cone and the plane surface, two unbounded curves are formed, which are mirror images of each other. A hyperbola is symmetric along the conjugate axis and resembles an ellipse in many ways. A hyperbola is a locus of points whose difference in the distances from two foci is a fixed value. This difference is obtained by subtracting the distance of the nearer focus from the distance of the farther focus. If P (x, y) is a point on the hyperbola and F, F’ are two foci, then the locus of the hyperbola is PF-PF’ = 2a.

### Hyperbola Equation

The standard equations of a hyperbola are,

(or)

A hyperbola has two standard equations. These equations of a hyperbola are based on its transverse axis and conjugate axis.

• The standard equation of the hyperbola is [(x2/a2) – (y2/b2)] = 1, where the X-axis is the transverse axis and the Y-axis is the conjugate axis.
• Furthermore, another standard equation of the hyperbola is [(y2/a2)- (x2/b2)] = 1, where the Y-axis is the transverse axis and the X-axis is the conjugate axis.
• The standard equation of the hyperbola with center (h, k) and the X-axis as the transverse axis and the Y-axis as the conjugate axis is,

• Furthermore, another standard equation of the hyperbola with center (h, k) and the Y-axis as the transverse axis and the X-axis as the conjugate axis is

### Derivation of Equation of the Hyperbola

Let us consider a point P on the hyperbola whose coordinates are (x, y). From the definition of the hyperbola, we know that the difference between the distance of point P from the two foci F and F’ is 2a, i.e., PF’-PF = 2a.

Let the coordinates of the foci be F (c, o) and F ‘(-c, 0).

Now, by using the coordinate distance formula, we can find the distance of point P (x, y) to the foci F (c, 0) and F ‘(-c, 0).

⇒ √[(x + c)2 + (y – 0)2] – √[(x – c)2 + (y – 0)2] = 2a

⇒ √[(x + c)2 + y2] = 2a + √[(x – c)2 + y2

Now, by squaring on both sides, we get

⇒ (x + c)2 + y2 = 4a2 + (x – c)2 + y2 + 4a√[(x – c)2 + y2

⇒ 4cx – 4a2 = 4a√[(x – c)2 + y2

⇒ cx – a2 = a√[(x – c)2 + y2

Now, by squaring on both sides and simplifying, we get

⇒ [(x2/a2) – (y2/(c2 – a2))] = 1

We have, c2 = a2 + b2, so by substituting this in the above equation, we get

⇒ x2/a2 – y2/b2 = 1

Hence, the standard equation of the hyperbola is derived.

Similarly, we can derive the standard equations of the other hyperbola, i.e., [y2/a2 – x2/b2] = 1.

### Terms used in Hyperbola

In analytic geometry, a hyperbola is a conic section that is developed when a plane cuts a double right circular cone at an angle such that both halves of the cone are joined. A hyperbola can be described using concepts like foci, directrix, latus rectum, and eccentricity.

Let us check through a few important terms relating to the different parameters of a hyperbola.

• Foci: A hyperbola has two foci whose coordinates are F(c, o), and F'(-c, 0).
• Center of a Hyperbola: The center of a hyperbola is the midpoint of the line that joins the two foci.
• Major Axis: The length of the major axis of a hyperbola is 2a units.
• Minor Axis: The length of the minor axis of a hyperbola is 2b units.
• Vertices: The points of intersection of the hyperbola with the axis are called the vertices. (a, 0), and (-a, 0) are the vertices of a hyperbola.
• Latus Rectum of Hyperbola: The latus rectum of a hyperbola is a line passing through any of the foci of a hyperbola and perpendicular to the transverse axis of the hyperbola. The endpoints of a latus rectum lie on the hyperbola, and its length is 2b2/a.

• Transverse Axis: The transverse axis of the hyperbola is the line that passes through the two foci and the center of the hyperbola.
• Conjugate Axis: The conjugate axis of the hyperbola is the line that passes through the center of the hyperbola and is perpendicular to the transverse axis.
• Asymptotes: A hyperbola has a pair of asymptotes, where an asymptote is a straight line that approaches the hyperbola on the graph, but never touches.

Equations of asymptotes of pair of asymptotes of a hyperbola: y = (b/a) x and y = -(b/a) x

• Directrix: The directrix of a hyperbola is a fixed straight line perpendicular to the axis of a hyperbola.
• Eccentricity of Hyperbola:  The eccentricity of a hyperbola is the ratio of the distance of a point from the focus to its perpendicular distance from the directrix. The eccentricity of a hyperbola is greater than 1, i.e., e >1.

Eccentricity of a hyperbola (e) = √[1 + (b2/a2)]

Hyperbola is an open curve that has two branches that look like mirror images of each other. For any point on any of the branches, the absolute difference between the point from foci is constant and equals 2a, where a is the distance of the branch from the center. The Hyperbola formula helps us to find various parameters and related parts of the hyperbola such as the equation of hyperbola, the major and minor axis, eccentricity, asymptotes, vertex, foci, and semi-latus rectum

### Sample Problems

Problem 1: Determine the eccentricity of the hyperbola x2/64 – y2/36 = 1.

Solution:

Given,

The equation of the hyperbola is x2/64 – y2/36 = 0

By comparing the given equation with the standard equation of the hyperbola x2/a2 – y2/b2 = 1, we get

a2 = 64, b2 = 36

⇒ a = 8, b = 6

We have,

Eccentricity of a hyperbola (e) = √(1 + b2/a2)

⇒ e = √(1 + 62/82)

⇒ e = √(1 + 36/64)

⇒ e = √(64 + 36)/64) = √(100/64)

⇒ e = 10/8 = 1.25‬

Hence, the eccentricity of the given hyperbola is 1.25‬.

Problem 2: If the equation of the hyperbola is [(x-4)2/25]-[(y-3)2/9] = 1, find the lengths of the major axis, minor axis, and latus rectum.

Solution:

Given,

The equation of the hyperbola is [(x-4)2/25]-[(y-3)2/9] = 1

By comparing the given equation with the standard equation of the hyperbola, (x – h)2/a2 – (y – k)2/b2 = 1

Here, x = 4 is the major axis and y = 3 is the minor axis.

a2 = 25 ⇒ a = 5

b2 = 9 ⇒ b = 3

The length of the major axis = 2a = 2 × (5) = 10 units

The length of the minor axis = 2b = 2 × (3) = 6 units

The length of the latus rectum = 2b2/a = 2(3)2/5 = 18/5 = 3.6 units

Problem 3: Find the vertex, asymptote, major axis, minor axis, and directrix if the hyperbola equation is [(x-6)2/72]-[(y-2)2/42] = 1.

Solution:

Given,

The equation of the hyperbola is [(x-6)2/72]-[(y-2)2/42] = 1

By comparing the given equation with the standard equation of the hyperbola, (x – h)2/a2 – (y – k)2/b2 = 1

h = 6, k = 2, a = 7, b = 4

The vertex of a hyperbola : (h + a, k) and (h – a, k) = (13, 2) and (-1, 2)

The major axis of the hyperbola is x = h ⇒ x = 6

The minor axis of the hyperbola is y = k ⇒ y = 2

The equations of asymptotes of the hyperbola are

y = k − (b / a)x + (b / a)h and y = k+ (b / a)x – (b / a)h

⇒ y = 2 – (4/7)x + (4/7)6 and y = 2 + (4/7)x – (4/7)6

⇒ y = 2 – 0.57x + 3.43 and y = 2 + 0.57x – 3.43

⇒ y = 5.43 – 0.57x and y = -1.43 + 0.57x

The equation of the directrix of a hyperbola is x = ± a2/√(a2 + b2)

⇒ x = ± 72/√(72 + 42) = ± 49/√65

⇒ x = ± 6.077

Problem 4: Find the eccentricity of the hyperbola whose latus rectum is half of its conjugate axis.

Solution:

Given,

The length of latus rectum is half of its conjugate axis.

Let the equation of hyperbola be [(x2 / a2) – (y2 / b2)] = 1

Then conjugate axis = 2b

The length of the latus rectum = (2b2 / a)

From the given data, (2b2 / a) = (1/2) × 2b

⇒ 2b = a

We have,

Eccentricity of a hyperbola (e) = √[1 + (b2/a2)]

Now, substitute a = 2b in the formula of eccentricity

⇒ e = √[1 + (b2/(2b)2]

⇒ e = √[1 + (b2/4b2)] = √(5/4)

⇒ e = √5/2

Hence, the required eccentricity is √5/2.

Problem 5: Find the vertex, foci, and equations of asymptotes if the hyperbola equation is [y2/25]-[x2/9] = 1.

Solution:

Given,

The equation of the hyperbola is [y2/25]-[x2/9] = 1 = 0

By comparing the given equation with the standard equation of the hyperbola y2/a2 – x2/b2 = 1, we get

a2 = 25, b2 = 9 ⇒ a = 5, b = 3

Coordinates of the vertex: (0, 5) and (0, -5)

Coordinates of foci: (0, c) and (0, -c)

We know that, c = √(a2 + b2) = √(52 + 32) = √34 = 5.83

Hence, coordinates of foci: (0, 5.83) and (0, -5.83)

Equations of asymptotes: y = (a/b) x and y = -(a/b) x

⇒ y = (5/3)x and y = -(5/3)x

Thus, equations of asymptotes are: 5x – 3y = and 5x + 3y = 0

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