Find the eccentricity, coordinates of the foci, equations of directrices and length of the latus-rectum of the hyperbola. (i) 4x^2 – 3y^2 = 36 (ii) 3x^2 – y^2 = 4
Find the eccentricity, coordinates of the foci, equations of directrices and length of the latus-rectum of the hyperbola. (i) 4x^2 – 3y^2 = 36 (ii) 3x^2 – y^2 = 4

(i)

    \[4{{x}^{2}}-\text{ }3{{y}^{2}}~=\text{ }36\]

Given:

The equation

    \[=>\text{ }4{{x}^{2}}-\text{ }3{{y}^{2}}~=\text{ }36\]

The equation can be expressed as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 17

The obtained equation is of the form

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 18

Where,

    \[{{a}^{2}}~=\text{ }9,\text{ }{{b}^{2}}~=\text{ }12\text{ }i.e.,\text{ }a\text{ }=\text{ }3\text{ }and\text{ }b\text{ }=\text{ }\surd 12\]

Eccentricity is given by:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 19

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 20

Foci: The coordinates of the foci are

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

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 21

    \[\left( \pm ae,\text{ }0 \right)\text{ }=\text{ }\left( \pm \surd 21,\text{ }0 \right)\]

The equation of directrices is given as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 22

The length of latus-rectum is given as:

    \[2{{b}^{2}}/a\]

    \[=\text{ }2\left( 12 \right)/3\]

Or,

    \[=\text{ }24/3\]

    \[=\text{ }8\]

(ii)

    \[3{{x}^{2}}-\text{ }{{y}^{2}}~=\text{ }4\]

Given:

The equation

    \[=>\text{ }3{{x}^{2}}-\text{ }{{y}^{2}}~=\text{ }4\]

The equation can be expressed as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 23

The obtained equation is of the form

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 24

Where,

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

Eccentricity is given by:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 25

Foci: The coordinates of the foci are

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

    \[\left( \pm ae,\text{ }0 \right)\text{ }=\text{ }\pm \left( 2/\surd 3 \right)\left( 2 \right)\text{ }=\text{ }\pm 4/\surd 3\]

    \[\left( \pm ae,\text{ }0 \right)\text{ }=\text{ }\left( \pm 4/\surd 3,\text{ }0 \right)\]

The equation of directrices is given as:

RD Sharma Solutions for Class 11 Maths Chapter 27 – Hyperbola - image 26

The length of latus-rectum is given as:

    \[2{{b}^{2}}/a\]

    \[=\text{ }2\left( 4 \right)/\left[ 2/\surd 3 \right]\]

So,

    \[=\text{ }4\surd 3\]