We will learn how to find the standard equation of a hyperbola.

Let S be the focus, e (> 1) the eccentricity and the line KZ its guideline for the hyperbola, the equation of which is required.

Standard equation of a hyperbola

From point S draw SK perpendicular to the guideline KZ. The route section SK and the generated SK share inside at A and outside at A ‘in the ratio e: 1.

Then,

( frac {SA} {AK} ) = e: 1

⇒ SA = e AK …………. (ii)

and ( frac {SA ‘} {A’K} ) = e: 1

⇒ SA ‘= e A’K …………………. (ii)


The points A and A ‘lie on the desired hyperbola, because according to the definition of the hyperbola, A and A’ are those points whose distance from the focal point has a constant ratio e (> 1) to their respective distance from the line, i.e. A and A ‘he on the required hyperbola.

Let AA ‘= 2a and let C be the center of the line segment AA’. Therefore CA = CA ‘= a.

Now draw CY perpendicular to AA ‘and mark the origin at C. CX and CY are assumed to be the x and y axes, respectively.

Now if we add the two equations (i) and (ii) above, we have

SA + SA ‘= e (AK + A’K)

⇒ CS – CA + CS + CA ‘= e (AC – CK + A’C + CK)

⇒ CS – CA + CS + CA ‘= e (AC – CK + A’C + CK)

Now set the value of CA = CA ‘= a.

⇒ CS – a + CS + a = e (a – CK + a + CK)

⇒2CS = e (2a)

⇒ 2CS = 2ae

⇒ CS = ae …………………… (iii)

Now we subtract the two above equations (i) from (ii) again and have:

⇒ SA ‘- SA = e (A’K – AK)

⇒ AA ‘= e {(CA’ + CK) – (CA – CK)}

⇒ AA ‘= e (CA’ + CK – CA + CK)

Now set the value of CA = CA ‘= a.

⇒ AA ‘= e (a + CK – a + CK)

⇒ 2a = e (2CK)

⇒ 2a = 2e (CK)

a = e (CK)

⇒ CK = ( frac {a} {e} ) ………………. (iv)

Let P (x, y) be any point on the required hyperbola and draw from P PM and PN perpendicular to KZ and KX, respectively. Join SP now.

According to the graphic, CN = x and PN = y apply.

Now form the definition of the hyperbola that we get,

SP = e PN

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

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

⇒ SP (^ {2} ) = e (^ {2} ) (CN – CK) (^ {2} )

⇒ (x – ae) (^ {2} ) + y (^ {2} ) = e (^ {2} ) (x – ( frac {a} {e} )) (^ {2} ), [From (iii) and (iv)]

⇒ x (^ {2} ) – 2aex + (ae) (^ {2} ) + y (^ {2} ) = (ex – a) (^ {2} )

⇒ (ex) (^ {2} ) – 2aex + a (^ {2} ) = x (^ {2} ) – 2aex + (ae) (^ {2} ) + y (^ {2} )

⇒ (ex) (^ {2} ) – x (^ {2} ) – y (^ {2} ) = (ae) (^ {2} ) – a (^ { 2} )

⇒ x (^ {2} ) (e (^ {2} ) – 1) – y (^ {2} ) = a (^ {2} ) (e (^ {2 }) – 1)

⇒ ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {a ^ {2} (e ^ {2} – 1)} ) = 1

We know that a (^ {2} ) (e (^ {2} ) – 1) = b (^ {2} )

Therefore ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1

The relationship applies to all points P (x, y) ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 fulfills the required hyperbola.

Hence the equation ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 represents the equation of the hyperbola.

The equation of a hyperbola in the form ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 is known as the standard equation of the hyperbola .

The hyperbole

11th and 12th grade math

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