If at any point (x_1, y_1) on the curve y=f(x | vedclass

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(B) We know that the length of the subtangent is given by $|y_1 \frac{dx}{dy}|$ and the length of the subnormal is given by $|y_1 \frac{dy}{dx}|$. Giv

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$\sqrt{2}|y_1|$

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(B) We know that the length of the subtangent is given by $|y_1 \frac{dx}{dy}|$ and the length of the subnormal is given by $|y_1 \frac{dy}{dx}|$. Given that the lengths of the subtangent and subnormal are equal at $(x_1, y_1)$, we have: $|y_1 \frac{dx}{dy}| = |y_1 \frac{dy}{dx}|$ Assuming $y_1 
eq 0$, we get: $|\frac{dx}{dy}| = |\frac{dy}{dx}|$ Since $\frac{dx}{dy} = \frac{1}{dy/dx}$, this implies: $|\frac{1}{dy/dx}| = |\frac{dy}{dx}|$ $|\frac{dy}{dx}|^2 = 1 \Rightarrow \frac{dy}{dx} = \pm 1$. The length of the tangent is given by the formula: $L = |y_1 \sqrt{1 + (\frac{dx}{dy})^2}|$ Since $\frac{dy}{dx} = \pm 1$, we have $\frac{dx}{dy} = \pm 1$. Substituting this into the formula: $L = |y_1 \sqrt{1 + (\pm 1)^2}| = |y_1 \sqrt{1 + 1}| = \sqrt{2}|y_1|$. Thus, the length of the tangent is $\sqrt{2}|y_1|$.

If at any point $(x_1, y_1)$ on the curve $y=f(x)$ the lengths of the subtangent and subnormal are equal, then the length of the tangent drawn to that curve at that point is

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