Conic Sections
Hyperbola Equilateral or rectangular hyperbola   Examples of hyperbola
Equilateral or rectangular hyperbola
Equilateral or rectangular hyperbola
The hyperbola whose semi-axes are equal, i.e., a = b
 has the equation x2 - y2 = a2.
Its asymptotes  y =  ± x are perpendicular and inclined to the x-axis at an angle of 45°.
Foci of the equilateral hyperbola,
F1(-Ö2 a, 0) and F2(Ö2 a, 0),
and the eccentricity  e = c/a = Ö2. Translated hyperbola
The equation of a hyperbola translated from standard position so that its center is at S(x0, y0)  is given by
b2(x - x0)2 - a2(y - y0)2 = a2b2
 or and after expanding and substituting constants obtained is
Ax2 + By2 + Cx + Dy + F = 0.
An equation of that form represents the hyperbola if
A · B < 0
that is, if coefficients of the square terms have different signs. Equation of the hyperbola in vertex form
By translating the hyperbola, centered at (0, 0), in the
negative direction of the
x-axis by x0 = -a, so that new
position of the center
S(-a, 0) then its equation is
b2(x + a)2 - a2y2 = a2b2.  After squaring and reducing,
 b2x2 + 2ab2x - a2y2 = 0 or since obtained is the equation of the hyperbola in vertex form. Parametric equation of the hyperbola
In the construction of the hyperbola, shown in the below figure, circles of radii a and b are intersected by an arbitrary line through the origin at points M and N. Tangents to the circles at M and N intersect the x-axis at R and S. On the perpendicular through S, to the x-axis, mark the line segment SP of length MR to get the point P of the hyperbola. We can prove that P is a point of the hyperbola.
In the right triangles ONS and OMR, by replacing OS = x and MR = SP
 and substituting by dividing by b2, therefore, P(x, y) is the point of the hyperbola. The coordinates of the point P(x, y) can also be expressed by the angle t common to both mentioned
 triangles, so that is the parametric equation of the hyperbola.
By substituting these parametrically expressed coordinates into equation of the hyperbola that is, known trigonometric identity.
Examples of hyperbola
Example:  Given is the hyperbola  4x2 - 9y2 = 36,  determine the semi-axes, equations of the asymptotes,
coordinates of foci, the eccentricity and the semi-latus rectum.
Solution:   Put the equation in the standard form to
determine the semi-axes, thus
4x2 - 9y2 = 36 | ¸ 36 Asymptotes, Applying, coordinates of foci,  F1(-Ö13, 0) and F2(Ö13, 0). The eccentricity, and the semi-latus rectum, Example:  Write equation of a hyperbola with the focus at F2(5, 0) and whose asymptotes are, Solution:  Therefore, the equation of the hyperbola, Example:  The hyperbola is given by equation  4x2 - 9y2 + 32x + 54y - 53 = 0.
Find coordinates of the center, the foci, the eccentricity and the asymptotes of the hyperbola.
Solution:   The given hyperbola is translated in the direction of the coordinate axes so the values of translations x0 and y0 we can find by using the method of completing the square rewriting the equation in
 the standard form, Thus,           4x2 + 32x - 9y2 + 54y - 53 = 0,
4(x2 + 8x) - 9(y2 - 6y) - 53 = 0
4[(x + 4)2 - 16] - 9[(y - 3)2 - 9] - 53 = 0
4(x + 4)2 - 9(y - 3)2 = 36 | ¸ 36
 Therefore,  it follows that   a2 = 9a = 3,   b2 = 4b = 2,  and the center of the hyperbola at  S(x0, y0)  or  S(-4, 3).
 Half the focal distance the eccentricity and the foci,  F1(x0 - c, 0)  so  F1(-4 - Ö13, 0)  and  F2(x0 + c, 0),   F2(-4 + Ö13, 0).
Equations of the asymptotes of a translated hyperbola therefore, the asymptotes of the given hyperbola, Example:   Write the equation of the hyperbola  9x2 - 25y2 = 225 in the vertex form.
Solution:   Using parallel shifting we should place the center of the hyperbola at S(-a, 0).
Rewrite    9x2 - 25y2 = 225 | ¸ 225 therefore,  a = 5 and  b = 3, so that S(-5, 0).
Then, the translated hyperbola with the center at  S(-5, 0) has the equation    Conic sections contents 