We will learn how to find the standard equation of a hyperbola.
Let S be the focus, e (> 1) be the eccentricity and line KZ its directrix of the hyperbola whose equation is required.
From the point S draw SK perpendicular to the directrix KZ. The line segment SK and the produced SK divides internally at A and externally at A’ respectively 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' he on the required hyperbola because according to the definition of hyperbola A and A’are such points that their distance from the focus bear constant ratio e (>1) to their respective distance from the directrix, therefore A and A' he on the required hyperbola.
Let AA’ = 2a and C be the midpoint 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 as x and yaxes respectively.
Now, adding the above two equations (i) and (ii) 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 put the value of CA = CA' = a.
⇒ CS  a + CS + a = e (a  CK + a + CK)
⇒2CS = e (2a)
⇒ 2CS = 2ae
⇒ CS = ae …………………… (iii)
Now, again subtracting above two equations (i) from (ii) we have,
⇒ SA'  SA = e (A'K  AK)
⇒ AA'= e {(CA’ + CK)  (CA  CK)}
⇒ AA' = e (CA’ + CK  CA + CK)
Now put 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 from P draw PM and PN perpendicular to KZ and KX respectively. Now join SP.
According to the graph, CN = x and PN = y.
Now form the definition of hyperbola we get,
SP = e ∙ PM
⇒ 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
For all the points P (x, y) the relation \(\frac{x^{2}}{a^{2}}\)  \(\frac{y^{2}}{b^{2}}\) = 1 satisfies on the required hyperbola.
Therefore, 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 of \(\frac{x^{2}}{a^{2}}\)  \(\frac{y^{2}}{b^{2}}\) = 1 is known as the standard equation of the hyperbola.
`● The Hyperbola
11 and 12 Grade Math
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