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Equation of a hyperbola in the standard form

x²/a² - y²/b² = 1

Let S be the focus, ZK be the directrix and e the eccentricity of the hyperbola whose equation is required. Draw SK perpendicular from S to the directrix ZK and divide SK internally and externally at A and A¹, A¹ being on SK, respectively in the ratio e:1.

Then SA = e.AK (1)

SA¹ = e.A¹K (2)

Since A and A¹ are such points that their distances from the focus bear a constant ratio e (>1) to their respective distances from the directrix; therefore, these points lie on the hyperbola.

Let C be the origin, CSX the x-axis and a straight line CY through C perpendicular to CX be the y-axis. Let P(x,y) be any point on the hyperbola. PM and PN are perpendicular from P onto ZK and KX respectively. By the definition

This is the equation of the hyperbola in the standard form.

Tracing a hyperbola

To trace the graph of the hyperbola

Symmetry:
For every value of x, there are equal and opposite values of y. Similarly, for every value of y there are equal and opposite values of x. Therefore, the curve is symmetric about both axes.
Origin:
The curve does not pass through the origin.
Intersection with the axes:
The curve meets the x axis at y = 0. If we put y = 0 in the equation (3), we get x = ± a. The curves meet the x-axis at A(a,0) A1 (-a,0). If we put x = 0 in equation (2), we get imaginary values for y. So the curve does not meet the y-axis.
Region:
From equation (2), we find that for -a< x < a the values of y are imaginary. So the curve does not exist between the lines x = -a and x = a. Also, y = 0 at x = ± a, and if x increases and is greater than a, the values of y also increase. Similarly, when it decreases and is less than -a, y also increases.

Second focus and second directrix of a hyperbola

The hyperbola x²/a² - y²/b² = 1 , b² = a²(e² -1) has a second focus S¹(-ae,0) and a second directrix Z¹K¹ with the equation x = -a/e.

Vertices, Transverse and Conjugate Axes, Foci, Directrices and Center of the Hyperbola:

x²/a² - y²/b² = 1

Vertices:

The points A, A¹ where the curve meets the line joining the foci S and S¹ are called the vertices of the hyperbola. The coordinates of A and A¹ are respectively (a,0) and (-a,0).

Transverse and Conjugate Axes:

The straight line joining the vertices A and A¹ is called the transverse axis of the hyperbola. Its length AA¹= 2a. The straight line through the center, which is perpendicular to the transverse axis, does not meet the hyperbola in real points. But if B, B¹ are points on this line such that CB = CB¹ = b, the line BB¹ is called the conjugate axis such that BB¹ = 2b.

Foci:

The points S(ae,0) and S¹(-ae,0) are called the foci of the hyperbola.

Directrices:

The lines ZK and Z¹K¹ are the two directrices of the hyperbola and their equations are x = a/e and x = -a/e, respectively.

Center:

The midpoint C of AA¹ bisects every chord of the hyperbola passing through it and is called the center of the hyperbola.

Eccentricity:

Length of latus rectum:

The hyperbola has the equation

x²/a² - y²/b² = 1

If LSL¹ is the latus rectum, LS is called the semi-latus rectum.

The coordinates of L are (ae,SL) as L lies on the parabola.

Focal Distances of a Point:

Asymptotes

Definition:

An asymptote is a straight line that meets the conic in two points, both of which are at infinity, but which itself is not altogether at infinity.

Find the asymptote to the hyperbola

x²/a² - y²/b² = 1 (1)

The asymptote is a straight line and, hence, can be

written as

The asymptotes pass through the center but never touch the curve, as given in the figure.

PP¹ and QQ¹ are the asymptotes.

Conjugate Hyperbola

The hyperbola whose transverse and conjugate axes are respectively the conjugate and transverse axes of a given hyperbola is called the conjugate hyperbola of the given hyperbola.

If the center of the hyperbola is at the point (h,k) and the direction of the axes are parallel to the coordinate axes, then its equation is

Example 2:

Find the equation of the hyperbola whose focus is (1, 2), directrix is 2x + y = 1 and eccentricity is √3.

Let S(1, 3) be the focus P(x,y) be a point on the hyperbola.

PM = perpendicular from P onto the directrix.

since A = 2 B = 1

For a hyperbola

SP/PM = e

is the required equation to the hyperbola.

Example 3:

For the following hyperbolas, find the lengths of the transverse and conjugate axis, the eccentricity, the coordinates of the foci, vertices, the length of the latus rectum, the equation of the directrices and the asymptotes.

16x² - 9y² = 144
3x² - 6y² = - 18

1. Solution:

Eccentricity:

Foci:

Vertices:

Coordinates are (a, 0), (-a, 0)

= (3, 0), (-3, 0)

Length of the latus rectum:

= 2b²/a

= 2 * 4²/3

= 2 * 16/3

= 32/3

Directrices:

Equation to the directrices are x = a/e, x = -a/e

= 3/(5/3) = -3/(5/3)

x = 9/5 x = -9/5

Asymptotes:

Equation to the asymptotes

2. Solution:

Foci:

Coordinates of the vertices are (0, be), (0, -be)

= (0, √3*√3), (0, -√3 * √3)

= (0, 3), (0, -3)

Vertices:

Coordinates of the vertices are (0, b), (0, -b)

= (0, √3), (0, -√3)

Length of the latus rectum:

Directrices:

Asymptotes:

Equation for the asymptotes

The graph of the above hyperbola is as below.

Example 3:

Show that the equation 9x² - 16y² - 18x + 32y - 151 = 0 represents a hyperbola. Find the coordinates of the center, foci, vertices, the eccentricity, the lengths of the latus recta, axes, the equation of the directrices and the asymptotes.

Solution:

Given equation 9x² - 16y² - 18x + 32y - 151 = 0

For brevity's sake, we will only prove

h²> ab

h = 0; a = 9; b = -16 (from the equation)

h² = 0

ab = -144

0 > -144

Shifting the origin to (1,1) and denoting the new coordinates with respect to the new axis as X,Y

x-1 = X

x = X + 1

y - 1 = Y

y = Y + 1

Equation (1) reduces to

Center:

The coordinates of the center with respect to the new axis is (0,0) with respect to the old axes

x = X + 1

= 0 + 1

x = 1

y = Y + 1

= 0 + 1

y = 1

Center is (1, 1)

Length of the transverse axis:

= 2a = 2 * 4 = 8

Length of the conjugate axis:

= 2b = 2*3 = 6

Eccentricity:

Foci:

With respect to the new axes, the coordinates of the foci are (ae, 0), (-ae, 0)

= (4 * 5/4, 0 ), (-4 * 5/4 , 0 )

= (5, 0 ), (-5, 0)

With respect to the old axes

x = X + 1

= 5 + 1

x = 6

y = Y + 1

= 0 + 1

y = 1

and

x = X + 1

= - 5 + 1

x = - 4

y = Y + 1

= 0 + 1

y = 1

Coordinates of the foci are (6, 1), (-4, 1)

Vertices:

With respect to the new axis, the vertices are (a, 0), (-a, 0)

= (4, 0), (-4, 0)

With respect to the old axis

x = X + 1

= 4 + 1

x = 5

y = Y + 1

= 0 + 1

y = 1

and

x = X + 1

= -4 + 1

x = -3

y = Y + 1

= 0 + 1

y = 1

Vertices are (5, 1), (-3, 1)

Length of the latus rectum:

Directrices:

Equation to the directrices with respect to the new axes are

x = a/e x = -a/e

= 4/5/4 =-4/5/4

X = 16/5 and X = -16/5

With respect to the old axes

x = X + 1

= 16/5 + 1

= 16+5/5

x = 21/5

5x = 21

or 5x - 21 = 0

x = X + 1

= -16/5 + 1

= -16 + 5 / 5

x = -11/5

5x = -11

or 5x + 11 = 0

are the equation of the directrices.

Asymptotes:

With respect to the new axes, the equations for the asymptotes are

Y = ± b/a X

Y = ± 3/4 X

With respect to the old axes

y - 1 = 3/4(x-1) and y - 1 = -3/4(x-1)

4y - 4 = 3x - 3 and 4y - 4 = -3x + 3

3x - 3 - 4y + 4 = 0 and 4y - 4 + 3x - 3 = 0

3x - 4y + 1 = 0 and 3x + 4y - 7 = 0

The graph of the above hyperbola is as below.

The hyperbola 9x² - 16y² - 18x + 32y - 151 = 0

Try these questions

1) Find the eccentricity, coordinates of the foci and vertices, equation of the directrices and asymptotes and length of latus rectum of the hyperbola. Give a rough sketch of the hyperbola.