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Question
Find the equation of the normal at a point on the curve x2 = 4y which passes through the point (1, 2). Also find the equation of the corresponding tangent.
Solution
Equations of the given circles are
`x^2+y^2=4 ................(1)`
`(x-2)^2+y^2=4...........(2)`
Equation (1) is a circle with centre O at the origin and radius 2. Equation (2) is a circle with centre C (2, 0) and radius 2. Solving equations (1) and (2), we have
`(x-2)^2+y^2=x^2+y^2`
`or x^2-4x+4+y^2=x^2+y^2`
or x=1 which gives `y=+-sqrt3`
Thus, the points of intersection of the given circles are `A(1,sqrt3) and A'(1,-sqrt3) ` as shown in the Figure
Required area of the enclosed region OACA′O between circles
= 2 [area of the region ODCAO]
= 2 [area of the region ODAO + area of the region DCAD]
`=2[int_0^1ydx+int_1^2ydx]`
`=2[int_0^1sqrt(4-(x-2)^2)dx+int_1^2sqrt(4-x^2)dx]`
`=2[1/2(x-2)sqrt(4-(x-2)^2)+1/2xx4sin^(-1)((x-2)/2)]_0^1+2[1/2xsqrt(4-x^2)+1/2xx4sin^(-1)(x/2)]_1^2`
`=[(x-2)sqrt(4-(x-2)^2)+4sin^(-1)((x-2)2)]_0^1+[xsqrt(4-x^2)+4sin^(-1)(x/2)]_1^2`
`=[(-sqrt3+4sin^(-1)(-1/2))-4sin^(-1)(-1)]+[4sin^(-1)1-sqrt3-4sin^(-1)(1/2)]`
`=[(-sqrt3-4xxpi/6)+4xxpi/2]+[4xxpi/2-sqrt3-4xxpi/6]`
`=(8pi)/3-2sqrt3`
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