Find the axes, eccentricity, latus-rectum and the coordinates of the foci of the hyperbola 25x^2 – 36y^2 = 225.
Find the axes, eccentricity, latus-rectum and the coordinates of the foci of the hyperbola 25x^2 – 36y^2 = 225.

Given:

The equation

    \[=>\text{ }25{{x}^{2}}-\text{ }36{{y}^{2}}~=\text{ }225\]

The equation can be expressed as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 33

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 34

The obtained equation is of the form

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 35

Where,

    \[a\text{ }=\text{ }3\text{ }and\text{ }b\text{ }=\text{ }5/2\]

Eccentricity is given by:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 36

Foci: The coordinates of the foci are

    \[\left( \pm ae,\text{ }0 \right)\]

    \[\left( \pm ae,\text{ }0 \right)\text{ }=\text{ }\pm 3\text{ }\left( \surd 61/6 \right)\text{ }=\text{ }\pm \text{ }\surd 61/2\]

    \[\left( \pm ae,\text{ }0 \right)\text{ }=\text{ }\left( \pm \text{ }\surd 61/2,\text{ }0 \right)\]

The equation of directrices is given as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 37

The length of latus-rectum is given as:

    \[2{{b}^{2}}/a\]

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 38

∴ Transverse

    \[axis\text{ }=\text{ }6,\text{ }conjugate\text{ }axis\text{ }=\text{ }5,\]

    \[e\text{ }=\text{ }\surd 61/6,\text{ }LR\text{ }=\text{ }25/6,\text{ }foci\text{ }=\text{ }\left( \pm \text{ }\surd 61/2,\text{ }0 \right)\]