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Vector Algebra

Question
CBSEENMA12032804

Using integration, find the area of the smaller region bounded by the curve straight x squared over 16 plus straight y squared over 9 space equals space 1 and the straight line straight x over 4 plus straight y over 3 equals 1.

Solution

The equation of ellipse is
                straight x squared over 16 plus straight y squared over 9 equals 1                          ...(1)
The equation of line is
               straight x over 4 plus straight y over 3 equals 1                               ...(2)

Let line (2) meet ellipse (1) in
A(4, 0) and B(0, 3)
Area of shaded region = Area of region OAB - area of increment space OAB
       equals space 3 over 4. integral subscript 0 superscript 4 square root of 16 minus straight x squared end root space dx space minus space 3 over 4 integral subscript 0 superscript 4 left parenthesis 4 minus straight x right parenthesis space dx    open square brackets because space space of space left parenthesis 1 right parenthesis comma space left parenthesis 2 right parenthesis close square brackets
         equals space 3 over 4 open square brackets fraction numerator straight x square root of 16 minus straight x squared end root over denominator 2 end fraction plus 16 over 2 sin to the power of negative 1 end exponent straight x over 4 close square brackets subscript 0 superscript 4 minus space 3 over 4 open square brackets fraction numerator left parenthesis 4 minus straight x right parenthesis squared over denominator left parenthesis negative 1 right parenthesis thin space left parenthesis 2 right parenthesis end fraction close square brackets subscript 0 superscript 4
equals space 3 over 4 open square brackets open parentheses 0 plus 8 cross times straight pi over 2 close parentheses minus left parenthesis 0 plus 0 right parenthesis close square brackets plus 6
equals space 3 straight pi space plus 6 space equals space 3 space left parenthesis straight pi plus 2 right parenthesis space square space units.

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