Find the area of the circle x 2 + y 2 = 16 which is exterior to the parabola y 2 = 6x. - Wired Faculty

Question

Find the area of the circle x² + y² = 16 which is exterior to the parabola y² = 6x.

Solution

The equation of circle is
   $straight x squared plus straight y squared space equals 16$ ...(1)
The equation of parabola is
   $straight y squared space equals 6 straight x$ ...(2)
From (1) and (2),
$straight x squared plus 6 straight x space equals space 16$
or $straight x squared plus 6 straight x minus 16 space equals space 0$
$therefore space space space left parenthesis straight x plus 8 right parenthesis space left parenthesis straight x minus 2 right parenthesis space equals space 0$
$rightwards double arrow$    $straight x equals negative 8 comma space 2$

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

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Vector Algebra

Find the area of the circle x² + y² = 16 which is exterior to the parabola y² = 6x.

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Vector Algebra

Find the area of the circle x2 + y2 = 16 which is exterior to the parabola y2 = 6x.

Some More Questions From Vector Algebra Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Find the area of the circle x² + y² = 16 which is exterior to the parabola y² = 6x.