The hyperbola \(\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} - \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1\) passes through the point \(\left( {3\sqrt 5 ,{\rm{\;}}1} \right)\) and the length of its latus rectum is \(\frac{4}{3}\) units. The length of the conjugate is

Concept:

Hyperbola: The locus of a point which moves such that its distance from a fixed point is greater than its distance from a fixed straight line. (Eccentricity = e > 1)

Calculation:

Given the hyperbola \(\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} - \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1\) passes through the point \(\left( {3\sqrt 5 ,{\rm{\;}}1} \right)\)

\(\Rightarrow \frac{{{{\left( {3\sqrt 5 } \right)}^2}}}{{{{\rm{a}}^2}}} - \frac{1}{{{{\rm{b}}^2}}} = 1\)

\(\Rightarrow \frac{{45}}{{{{\rm{a}}^2}}} - \frac{1}{{{{\rm{b}}^2}}} = 1\)       ---(1)

Now length of latus rectum = \(\frac{{2{{\rm{b}}^2}}}{{\rm{a}}}\)

\(\Rightarrow \frac{{2{{\rm{b}}^2}}}{{\rm{a}}} = \frac{4}{3}\)

\(\therefore {\rm{a}} = \frac{{3{{\rm{b}}^2}}}{2}\)       ---(2)

Putting the value of 'a' in equation (1),

\(\Rightarrow \frac{{45}}{{{{\left( {\frac{{3{{\rm{b}}^2}}}{2}} \right)}^2}}} - \frac{1}{{{{\rm{b}}^2}}} = 1\)

\(\Rightarrow \frac{{20}}{{{{\rm{b}}^4}}} - \frac{1}{{{{\rm{b}}^2}}} = 1\)

⇒ 20 - b² = b⁴

⇒ b⁴ + b² - 20 = 0

⇒ b⁴ + 5b² - 4b² - 20 = 0

⇒ b² (b² + 5) - 4(b² + 5) = 0

⇒ (b² - 4) (b² + 5) = 0

⇒ (b² - 4) = 0

⇒ b² = 4

∴ b = 2