-->

Vector Algebra

Question
CBSEENMA12032819

The area of the circle x +y =16 exterior to the parabola y2 = 6x is
  • 4 over 3 left parenthesis 4 straight pi minus square root of 3 right parenthesis
  • 4 over 3 left parenthesis 4 straight pi plus square root of 3 right parenthesis
  • 4 over 3 left parenthesis 8 straight pi minus square root of 3 right parenthesis
  • 4 over 3 left parenthesis 8 straight pi plus square root of 3 right parenthesis

Solution

C.

4 over 3 left parenthesis 8 straight pi minus square root of 3 right parenthesis

The equation of circle is
                     straight x squared plus straight y squared space equals space 16                     ...(1)
The equation of parabola is
                  straight y squared space equals space 6 straight x                                  ...(2)
From (1) and (2),
                straight x squared plus 6 straight x space equals space 16
or      straight x squared plus 6 straight x minus 16 space equals space 0
therefore space space left parenthesis straight x plus 8 right parenthesis space left parenthesis straight x minus 2 right parenthesis space equals space 0
rightwards double arrow                       x = -8, 2
Rejecting x = -8 as parabola lies in 1st or 4th quadrant, we get x = 2
When              x = 2,   straight y equals square root of 12 space equals space 2 square root of 3
therefore   (1) and (2) intersect in straight P left parenthesis 2 comma space 2 square root of 3 right parenthesis
Required Area = Area of the circle - area of circle interior to the parabola
                        equals space 16 straight pi minus 2 space integral subscript 0 superscript 2 straight y space dx space minus space 2 integral subscript 2 superscript 4 straight y space dx
equals space 16 straight pi minus 2 integral subscript 0 superscript 2 square root of 6 space straight x to the power of 1 half end exponent dx space minus space 2 integral subscript 0 superscript 4 square root of 16 minus straight x squared end root space dx
space equals space 16 straight pi space minus space 2 square root of 6 open square brackets 2 over 3 straight x to the power of 3 over 2 end exponent close square brackets subscript 0 superscript 2 space minus space 2 open square brackets straight x over 2 square root of 16 minus straight x squared end root plus 16 over 2 sin to the power of negative 1 end exponent straight x over 4 close square brackets subscript 2 superscript 4
equals 16 straight pi minus 4 over 3 square root of 6. space 2 square root of 2 minus 2 open square brackets 4 over 2 square root of 16 minus 6 end root plus 16 over 2 sin to the power of negative 1 end exponent 1 minus 2 over 2 square root of 16 minus 4 end root minus 16 over 2 sin to the power of negative 1 end exponent 2 over 4 close square brackets
equals space 16 straight pi minus fraction numerator 16 over denominator square root of 3 end fraction minus 16. space straight pi over 2 plus 4 square root of 3 plus fraction numerator 8 straight pi over denominator 3 end fraction
equals negative fraction numerator 4 over denominator square root of 3 end fraction plus fraction numerator 32 straight pi over denominator 3 end fraction equals space 4 over 3 left parenthesis 8 straight pi minus square root of 3 right parenthesis

Some More Questions From Vector Algebra Chapter