We will discuss the definition of hyperbola and how to find the equation of the hyperbola given its focus, guideline and eccentricity.

If a point (P) moves in the plane in such a way that the ratio of its distance from a fixed point (S is called focus) in the same plane to its perpendicular distance from the fixed line (L is called dirctrix) is always constant, which is always greater than one , then the location drawn by P is called hyperbola.


Let S and L be a fixed point or a fixed straight line on a plane. If point P moves on this plane in such a way that its distance from fixed point S is always in a constant ratio to its perpendicular distance from fixed line L and its ratio is greater than one, then the location of point P is called a hyperbola.

Definition of hyperbola

The fixed point S is called the focal point and the fixed line L, the associated guideline and the constant ratio of the eccentricity of the hyperbola. The eccentricity is generally referred to as e (> 1).

If S is the focus, Z is the guideline, and P is any point on the hyperbola then is by definition

( frac {SP} {PM} ) = e

⇒ SP = e PN

Solved example to find the equation of Hyperbola, the focus, directrix and eccentricity of which are given:

The equation of the guideline of a hyperbola is x + y = -1. Its focus is on (1, 2) and the eccentricity is ( frac {3} {2} ). Find the equation of the hyperbola.

Solution:

Let P (x, y) be an arbitrary point on the required hyperbola. If PM is the length of the normal of P on the guideline x + y = -1 or x + y + 1 = 0 then

PM = ( frac {x + y + 1} { sqrt {1 ^ {2} + (-1) ^ {2}}} ) = ( frac {x + y + 1} {√2 } )

Here, too, the distance from P to the focus S is (- 1, 1)

SP = ( sqrt {(x – 1) ^ 2 + (y – 1) ^ 2} )

Since the point Ply lies on the required hyperbola, so by definition we have

( frac {SP} {PM} ) = e

SP = e PN

⇒ SP (^ {2} ) = e (^ {2} ) (PM) (^ {2} )

⇒ (x – 1) (^ {2} ) + (y – 2) (^ {2} ) = ( frac {9} {4} ) ( frac {(x + y + 1) ^ {2}} {2} ), [Since, e = 3]

⇒ 8x (^ {2} ) + 8y (^ {2} ) – 16x – 32y + 40 = 9x (^ {2} ) + 9y (^ {2} ) + 9 + 18xy + 18x + 18y

⇒ x (^ {2} ) + y (^ {2} ) + 18xy + 34x + 50y – 31 = 0, which is the required equation of the hyperbola.

The hyperbole

11th and 12th grade math degree

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