Binomial Theorem

Question
CBSEENMA11015165

In the following equation, Find the co-ordinates of foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of latus rectum of the ellipse.
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Solution

Equation of ellipse is : <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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rightwards double arrow               fraction numerator 4 straight x squared over denominator 36 end fraction plus fraction numerator 9 straight y squared over denominator 36 end fraction equals 1 space space space rightwards double arrow space space straight x squared over 9 plus straight y squared over 4 equals 1
Now, 9 greater than 4 space rightwards double arrow space straight a squared equals 9 comma space straight b squared equals 4 space rightwards double arrow space straight a equals 3 comma space straight b equals 2
∴    Equation of ellipse in the standard form is space space space straight x squared over straight a squared plus straight y squared over straight b squared equals 1
(i) Focus,      straight S subscript 1 left parenthesis negative straight c comma space 0 right parenthesis space left right arrow space open parentheses negative square root of straight a squared minus straight b squared end root comma space 0 close parentheses space left right arrow space left parenthesis negative square root of 9 minus 4 end root comma space 0 right parenthesis space left right arrow space left parenthesis negative square root of 5 comma space 0 right parenthesis
Focus,               straight S subscript 2 space left parenthesis straight c comma space 0 right parenthesis space left right arrow space left parenthesis square root of 5 comma space 0 right parenthesis
(ii) Vertex,           straight A space left parenthesis straight a comma space 0 right parenthesis space left right arrow space left parenthesis 3 comma space 0 right parenthesis
Vertex,            straight A apostrophe left parenthesis negative straight a comma space 0 right parenthesis space left right arrow space left parenthesis negative 3 comma space 0 right parenthesis
(iii) Length of major axis = 2a = 2(3) = 6
      Length of minor axis = 2b = 2(2) = 4
(iv)  Let the eccentricity be e.
Now,                           c = ae,  rightwards double arrow  square root of 5 space equals space 3 straight e   ∴   straight e equals fraction numerator square root of 5 over denominator 3 end fraction
(v) Length of latus rectum equals space fraction numerator 2 straight b squared over denominator straight a end fraction space equals space fraction numerator 2 left parenthesis 4 right parenthesis over denominator 3 end fraction equals 8 over 3

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