Hyperbola Formula

HYPERBOLA FORMULA

In a simple sense, hyperbolas resemble mirrored parabolas. The branches are the two halves. A parabola is formed when the plane intersects the halves of a right circular cone at an angle parallel to the cone’s axis.

Hyperbolas have two foci and two vertices. A hyperbola’s foci are away from its centre and vertices. A Hyperbola Formula is a conic section formed by the intersection of a double cone and a plane surface, but not necessarily at its centre. Like the ellipse, a Hyperbola Formula is symmetric along the conjugate axis. Hyperbolas have foci, directrix, latus rectum, and eccentricity. Examples of hyperbolas include the path taken by the shadow of a sundial, the scattering trajectory of subatomic particles, etc.

The purpose of this chapter is to provide an understanding of the hyperbola definition, formula, derivation of the formula, and standard forms of a Hyperbola Formula through the solving of examples.

What is Hyperbola?

Hyperbolas are smooth curves that lie on a plane and have two components or branches that are mirror images of each other and resemble infinite bows. Hyperbolas are sets of points whose distances from two foci are constant. It is calculated by subtracting the distance from the farther focus from the distance from the closer focus. The locus of the Hyperbola Formula is PF – PF’ = 2a for a point P(x,y) on the hyperbola and two foci F, F’.

Hyperbola Definition

In analytic geometry, a Hyperbola Formula is formed when a plane intersects a double right circular cone at an angle such that both halves are intersected. A hyperbola is created when the plane and cone intersect, resulting in two unbounded curves that are mirror images of one another.

Parts of a Hyperbola

The Hyperbola Formula has two foci with coordinates F(c, o) and F'(-c, 0).

Centre of hyperbola is the centre point of the line connecting the two foci.

A hyperbola’s major axis measures 2a units in length.

Any hyperbola’s minor axis measures 2b units length wise.

Hyperbola Formula vertices are the points where the axis intersects the hyperbola. (a, 0), (-a, 0) are the vertices of a hyperbola.

The latus rectum of a hyperbola passes through the foci of the hyperbola and is drawn perpendicular to its transverse axis. The length of the latus rectum of Hyperbola Formula is 2b2/a.

The transverse axis of the hyperbola passes through the two foci and the centre.

Hyperbola Formula conjugate axis: This line perpendicular to the transverse axis passes through the centre of the hyperbola.

Eccentricity of Hyperbola: (e > 1) Eccentricity is the ratio between the distance of the focus from the centre of the Hyperbola Formula, and the distance of the vertex from the centre. Focus distance is ‘c’ units, and vertex distance is ‘a’ units, so the eccentricity equals e = c/a.

Hyperbola Equation

A hyperbola can be represented by the following equation. The x-axis represents the transverse axis of the Hyperbola Formula, and the y-axis represents its conjugate axis.

x2/a2-y2/b2=1

Standard Equation of Hyperbola

The Hyperbola has two standard equations. Each hyperbola has a transverse axis and conjugate axis. The standard equation of hyperbola is x2/a2-y2/b2=1 has the transverse axis as the x-axis and the conjugate axis as y-axis. Further, another equation of Hyperbola Formula is y2/a2-x2/b2=1, and it has the transverse axis as y-axis and its conjugate axis is x-axis.

Derivation of Hyperbola Equation

P(x, y) should be on the hyperbola such that PF1 – PF2 = 2a

As a result of the distance formula, we have:

√ {(x + c)2 + y2} – √ {(x – c)2 + y2} = 2a

Or, √ {(x + c)2 + y2} = 2a + √ {(x – c)2 + y2}

On squaring both sides as well. As a result,

(x + c)2 + y2 = 4a2 + 4a√ {(x – c)2 + y2} + (x – c)2 + y2

As a result of simplifying the equation,

√ {(x – c)2 + y2} = x(c/a) – a

Squaring both sides again and simplifying further,

x2/a2 – y2/(c2 – a2) = 1

It is known that c2 – a2 = b2. As a result,

x2/a2 – y2/b2 = 1

As a result, any point on the Hyperbola Formula satisfies the equation:

x2/a2 – y2/b2 = 1

Hyperbola Formula

Graph of Hyperbola

• Axis minor

Perpendicular to the major axis, it crosses the center of the hyperbola.

The minor axis has a length of 2b. Therefore, the equation is as follows:

x = x0

• Axis major

A major axis is the line that crosses by the middle, the focus of the Hyperbola Formula, and the vertices. 2a is the length of the major axis. Here is the equation:

y = y0

• Asymptotes

Asymptotes are two intersecting line segments that cross through the centre of the Hyperbola Formula without touching the curve. Asymptotes can be expressed as follows:

y = y0 + b/ax – b/ax0

y = y0 − b/ax + b/ax0

Each hyperbola consists of two curves, each with a vertex and a focus. A hyperbola’s transverse axis crosses both its vertices and foci, and its conjugate axis is perpendicular to it. If the equation of Hyperbola Formula is not in standard form, then one needs to complete the square to get standard form.

The Extramarks website provides more information about Hyperbola graphs.

Properties of a Hyperbola

Hyperbola can be better understood by considering the following properties related to different concepts.

Asymptotes are straight lines drawn parallel to the Hyperbola Formula and assumed to touch the hyperbola at infinity. Asymptotes of the hyperbola have equations y = bx/a and y = -bx/a.

A hyperbola with the same transverse and conjugate axes is called a rectangular hyperbola. In this case, 2a = 2b, or a = b. Hence, equation of rectangular Hyperbola Formula is equal to x2 – y2 = a2

Hyperbola points can be represented by the parametric coordinates (x, y) = (asecθ, btanθ). The parametric coordinates of these points on the hyperbola satisfy its equation.

The auxiliary circle is drawn using the endpoints of the transverse axis of the hyperbola as its diameter. The equation of the auxiliary circle of Hyperbola Formula is x2 + y2 = a2.

The director circle is the point of intersection of perpendicular tangents to the hyperbola. The equation of the director circle is x2 + y2 = a2 – b2.

Examples on Hyperbola

1: Find the hyperbola’s equation if e1 is the eccentricity of the ellipse, x2/16 + y2/25 = 1, and e2 is the eccentricity of the Hyperbola Formula passing through the foci of the ellipse, e1e2 = 1.

Solution: The eccentricity of x2/16 + y2/25 = 1 is

e1 = √(1-16/25) = 3/5

e2 = 5/3

This is obtained using the relation e1e2 = 1.

Hence, the foci of the ellipse are (0, ± 3)

Hence, the equation of the Hyperbola Formula is x2/16 – y2/9 = -1.

2: A hyperbola, having transverse axis of length 2 sin θ, is confocal with ellipse 3×2 + 4y2 = 12. Then its equation is

1. x2cosec2 θ – y2sec2 θ = 1                2. x2sec2 θ – y2cosec2θ = 1
2. x2sin2 θ – y2cos2 θ = 1                   4. x2cos2 θ – y2sin2 θ = 1

Solution: Given ellipse is x2/4 + y2/3 = 1

Hence, a = 2 and b = √3

Hence, 3 = 4(1-e2) which gives e = 1/2

So, ae = 2.1/2 = 1

Thus, the eccentricity e1 for the Hyperbola Formula is

1 = e1 sin θ which means e1 = cosec θ.

So, b2 = sin2 θ(cosec2 θ – 1) = cos2 θ

Hence, equation of hyperbola is x2/sin2 θ – y2/cos2θ = 1

Or x2cosec2 θ – y2sec2 θ = 1

3: The circle whose equation is x2 + y2 – 8x = 0 and Hyperbola Formula x2/9 – y2/4 = 1 intersect at the points A and B.

An equation for a common tangent to the circle and hyperbola with a positive slope is…?

Find the equation of the circle which has a diameter AB.

Solution: (1) Equation of the tangent to the Hyperbola Formula having slope m is y = mx + √9m2-4

The equation of the tangent to the circle is y = m(x-4) + √16m2 +16

The equations will be identical when m = 2/√5

Therefore, the equation of common tangent is 2x – √5y + 4 = 0

(2) The equation of the Hyperbola Formula is x2/9 – y2/4 = 1 and that of circle is x2 + y2 – 8x = 0.

For their points of intersection, x2/9 + (x2– 8x)/ 4 = 1

So, this gives 4×2 + 9×2 -72x = 36

So, 13×2 -72x -36 = 0

This gives x = 6 and -13/6

But x = -13/6 is not acceptable

Now, x = 6, y = ± 2√3

Required equation is (x-6)2 + (y+2√3)(y-2√3) = 0

This gives x2 + y2 -12x + 24 = 0

4: Find equation of the tangent to the Hyperbola Formula having equation x2/9−y2=1 whose slope is 5

Solution:

Slope of tangent m = 5, a2 = 9, b2 = 1

Tangent equation in slope form is

y = mx ± √(a2m2 – b2 )

y = 5x ± √(9.52 – 1)

y = 5x ± 4√14

[Note: For the ellipse, director circle is x2 + y2 = a2 + b2, x2/a2 + y2/b2 = 1]

Normal:

Equation of the normal of x2/a2 – y2/b2 = 1 at (x1, y1)

a2x/x1 + b2y/y1 = a2 + b2

5: Find the normal at point (6, 3) to the Hyperbola Formula having equation x2/18 − y2/9 = 1

Solution:

Equation of the Normal at point (x1, y1) is a2 = 18, b2 = 9

a2x/x1 + b2y/y1 = a2 + b2

At point (6, 3), the normal’s equation is

18x/6 + 9y/3 = 18 + 9

x + y = 9

6: Find the equation of the chord of Contact of the point (2, 3) to the Hyperbola Formula x2/16 − y2/9 = 1

Solution:

T = 0 is the equation of the chord of contact

i.e. (xx1)/a2 – (yy1)/b2 – 1 = 0

Or, 2x/16 – 3y/9 = 1

Or, x/8 – y/3 = 1

A chord’s equation when its mid-point is known

T = (xx1)/a2 – (yy1)/b2 – 1 = x12/a2 – y12/b2 − 1.

7: Find equation of the chord of the hyperbola x2/9 – y2/4 = 1 whose midpoint is (5, 1).

Solution: Equation of chord of the Hyperbola Formula whose mid-points is (x1, y1)

T = (xx1)/a2 – (yy1)/b2 – 1 = x12/a2 – y12/b2 − 1

Given, midpoint is (5, 1).

=>5x/9 – y/4 – 1= 25/9 – ¼ – 1

=>5x/9 – y/4 = 91/36

Practice Questions on Hyperbola

1. Find the equation of parabola with focus (− √2, 0) and directrix x = √2.

Solution

The parabola is open left and has the x-axis and vertex (0,0) as its axis of symmetry.

Therefore, the equation of the parabola which is required is

(y – 0)2 = – 4√2 ( x – 0)

⇔ y2 = – 4√2x.

2. This parabola has a vertex of (5, -2) and a focus of (2, -2). Find the equation of this parabola.

Solution

The focal distance AS = a = 3 is measured from vertex A (5, -2), focus S (2, -2), and focal distance AS = b = 5.

The parabola is open on the left and symmetric about the x-axis.

Therefore, the equation of the parabola which is required is

(y + 2)2 = -4 (3)(x – 5)

⇔ y2 + 4 y + 4 = -12x + 60

⇔ y2 + 4 y +12x – 56 = 0.

3. Determine the equation of the parabola with vertex (-1, -2) and axis parallel to the y-axis and passing through (3, 6).

Solution

Since the axis is parallel to the y-axis, the parabola equation is:

(x +1)2 = 4a (y + 2).

As this passes through (3,6), we get

(3 +1)2 = 4a (6 + 2)

⇔ a = 1/2.

Then the equation of parabola is (x +1)2 = 2 (y + 2) which on simplifying yields,

x2 + 2x – 2y – 3 = 0.

4. In the parabola x2 – 4x – 5 y -1 = 0, find the vertex, focus, directrix, and length of the latus rectum.

Solution

For the parabola,

x2 – 4x – 5y -1 = 0

⇔ x2 – 4x = 5y +1

⇔ x2 – 4x + 4 = 5y +1+ 4.

⇒ (x – 2)2 = 5(y +1) is in standard form. Therefore, 4a = 5 the focus is (2, 1/4) and the vertex is (2, -1).

Directrix equation is y + k + a = 0

y -1+ 5/4 = 0

4y +1 = 0.

The length of the latus rectum is five units.

5. Find the equation of ellipse with foci (± 2, 0), vertices (± 3, 0).

Solution

SS′  =  2c and 2c = 4; A′A = 2a = 6

⇒ a = 3 and c  =  2,

⇒ b2 = a2 − c2 = 9 − 4 = 5.

The major axis runs along the x-axis, since a > b.

Centre is (0, 0) and Foci are (±2, 0).

Thus, x 2/9 + y 2/5 = 1 is the equation of the ellipse