If the length of the subnormal is equal to th | vedclass

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(D) The length of the subtangent is given by $|y / (dy/dx)|$ and the length of the subnormal is given by $|y \cdot (dy/dx)|$.Given that the length of

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$\frac{49}{2}$

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(D) The length of the subtangent is given by $|y / (dy/dx)|$ and the length of the subnormal is given by $|y \cdot (dy/dx)|$.Given that the length of the subnormal equals the length of the subtangent,we have $|y \cdot (dy/dx)| = |y / (dy/dx)|$,which implies $(dy/dx)^2 = 1$,so $dy/dx = \pm 1$.Case $1$: If $dy/dx = 1$,the tangent at $(3, 4)$ is $y - 4 = 1(x - 3)$,which simplifies to $y = x + 1$. This line intersects the axes at $(0, 1)$ and $(-1, 0)$. Since the problem specifies the positive coordinate axes,this case is rejected.Case $2$: If $dy/dx = -1$,the tangent at $(3, 4)$ is $y - 4 = -1(x - 3)$,which simplifies to $x + y = 7$. This line intersects the positive coordinate axes at $A(7, 0)$ and $B(0, 7)$.The area of $\Delta OAB$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 7 	imes 7 = \frac{49}{2}$.

If the length of the subnormal is equal to the length of the subtangent at any point on the curve $y = f(x)$ and the tangent at $(3, 4)$ to $y = f(x)$ meets the positive coordinate axes at $A$ and $B$,then the area of $\Delta OAB$,where $O$ is the origin,is

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