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Using integration find the area of the triangle formed by positive x-axis and tangent and normal to the circle x2 + y2 = 4at (1, √3).

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Given circle is x2 + y2 = 4

⇒ 2x + 2y \(\frac{dy}{dx}\) = 0 [By differentiating]

⇒ \(\frac{dy}{dx}=-\frac{x}{y}\)

Now, slope of tangent at (1, √3) = \(\frac{dy}{dx}]_{(1,\sqrt3)}=-\frac{1}{\sqrt3}.\)

\(\therefore\) Slope of normal at (1, √3) = √3

Therefore, equation of tangent is

\(\frac{y-\sqrt3}{x-1}=-\frac{1}{\sqrt3}\)

⇒ x + √3y = 4.....(i)

Again, equation of normal is

\(\frac{y-\sqrt3}{x-1}=\sqrt3\)

⇒ y - √3x = 0....(ii)

To draw the graph of the triangle formed by the lines x-axis, (i) and (ii), we find the intersecting points of these three lines which give vertices of required triangle. Let O, A, B be the intersecting points of these lines.

Obviously, the coordinate of O, A, B are (0, 0), (1, √3) and (4, 0) respectively.

Required area = area of triangle OAB = area of region OAC + area of region CAB

\(=\int_0^1y\,dx+\int_0^4\,dx\)

[Where in 1st integrand y = √3x and in 2nd y = \(\frac{4-x}{\sqrt3}\)]

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