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(D) Given curve is $x^2+y^2=16a^2$. On differentiating the equation with respect to $x$,we get $2x + 2yy' = 0$,which implies $y' = -\frac{x}{y}$. At t

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$8a^2$

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(D) Given curve is $x^2+y^2=16a^2$. On differentiating the equation with respect to $x$,we get $2x + 2yy' = 0$,which implies $y' = -\frac{x}{y}$. At the point $(2\sqrt{2}a, 2\sqrt{2}a)$,the slope of the tangent is $m_1 = -\frac{2\sqrt{2}a}{2\sqrt{2}a} = -1$. The slope of the normal is $m_2 = -\frac{1}{m_1} = 1$. The equation of the tangent is $y - 2\sqrt{2}a = -1(x - 2\sqrt{2}a)$,which simplifies to $x + y = 4\sqrt{2}a$. The $X$-intercept of the tangent is found by setting $y=0$,giving $x = 4\sqrt{2}a$. Let this point be $B(4\sqrt{2}a, 0)$. The equation of the normal is $y - 2\sqrt{2}a = 1(x - 2\sqrt{2}a)$,which simplifies to $y = x$. The normal passes through the origin $O(0,0)$ and the point $A(2\sqrt{2}a, 2\sqrt{2}a)$. The triangle is formed by the points $O(0,0)$,$A(2\sqrt{2}a, 2\sqrt{2}a)$,and $B(4\sqrt{2}a, 0)$. The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$. Area $= \frac{1}{2} |0(2\sqrt{2}a-0) + 2\sqrt{2}a(0-0) + 4\sqrt{2}a(0-2\sqrt{2}a)| = \frac{1}{2} |4\sqrt{2}a(-2\sqrt{2}a)| = \frac{1}{2} |-16a^2| = 8a^2$.

The area of the triangle formed by the positive $X$-axis,the tangent and the normal to the curve $x^2+y^2=16a^2$ at the point $(2\sqrt{2}a, 2\sqrt{2}a)$ is

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