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Hyperbola – Analytic geometry Formulas

Hyperbola

In analytic geometry a hyperbola is a conic section formed by intersecting a right circular cone with a plane at an angle such that both halves of the cone are intersected. This intersection produces two separate unbounded curves that are mirror images of each other.

Hyperbola with center at the origin:

Hyperbola with center at the origin and transverse axis on the x-axis.

x2a2–y2b2=1 \frac {x^2}{a^2} – \frac {y^2}{b^2} = 1

Hyperbola with center at the origin and transverse axis on the y-axis

y2a2–x2b2=1 \frac {y^2}{a^2} – \frac {x^2}{b^2} = 1

Example:

If the latus rectum of a hyperbola is 8 and eccentricity be 3 / √5, then what is the equation of the hyperbola?

2b2 / a = 8 and 3/√5 = √(1 + b2) / a2

4 / 5 = b2 / a2

a = 5, b = 2√5

Hence, the required equation of the hyperbola is (x2 / 25) − (y2 / 20) = 1

4x2 − 5y2 = 100

Hyperbola with center at any point (h, k):

Hyperbola with center at (h, k) and transverse axis parallel to the x-axis.

(x–h)2a2–(y–k)2b2=1 \frac {(x-h)^2}{a^2} – \frac {(y-k)^2}{b^2} = 1

Hyperbola with center at (h, k) and transverse axis parallel to the y-axis.

(y–k)2a2–(x–h)2b2=1 \frac {(y-k)^2}{a^2} – \frac {(x-h)^2}{b^2} = 1

Elements of Hyperbola:

    • Center (h, k). At the origin, (h, k) is (0, 0).
    • Transverse axis = 2a and conjugate axis = 2b
    • Location of foci c, relative to the center of hyperbola.
c=a2+b2 c=\sqrt{a^2+b^2}
    • Latus rectum, LR
2b2a 2 \frac {b^2}{a}
    • Eccentricity, e
e=ca>1.0 e= \frac ca \gt 1.0
    • Location of directrix d relative to the center of hyperbola.
d=ae ord=a2c d= \frac ae \ or d= \frac {a^2}{c}
    • Equation of asymptotes.
y–k=±m(x–h) y-k= \pm m(x-h)

where,

m is (+) for upward asymptote and m is (-) for downward.

m = b/a if the transverse axis is horizontal

m = a/b if the transverse axis is vertical

Example:

Find the transverse axis, the center, the foci and the vertices of the hyperbola whose equation is

16 y2 – x2 =16

Divide all terms of the given equation by 16 which becomes y2 – x2 / 16 = 1

Transverse axis: y axis or x = 0

center at (0 , 0)

vertices at (0 , 1) and (0 , -1)

c2 = 1 + 16 = 17. Foci are at (0 , √17) and (0 , -√17).

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