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Question

(B) Let the curve be $y = f(x)$. The length of the subtangent is given by $|y_1 / (dy/dx)|$ and the length of the subnormal is given by $|y_1 (dy/dx)|

Vedclass · Accepted Answer

$\sqrt{2} y_1$

Vedclass · Answer

(B) Let the curve be $y = f(x)$. The length of the subtangent is given by $|y_1 / (dy/dx)|$ and the length of the subnormal is given by $|y_1 (dy/dx)|$.Given that the length of the subtangent equals the length of the subnormal:$|y_1 / (dy/dx)| = |y_1 (dy/dx)|$Assuming $y_1 
eq 0$,we have $1 / |dy/dx| = |dy/dx|$,which implies $(dy/dx)^2 = 1$,so $dy/dx = \pm 1$.The length of the tangent is given by the formula $L_t = |y_1| \sqrt{1 + (dx/dy)^2}$.Since $dy/dx = \pm 1$,then $dx/dy = \pm 1$.Substituting this into the formula:$L_t = |y_1| \sqrt{1 + (\pm 1)^2} = |y_1| \sqrt{1 + 1} = \sqrt{2} |y_1|$.Thus,the length of the tangent is $\sqrt{2} y_1$ (assuming $y_1 > 0$).

If the length of the subtangent and the length of the subnormal at any point $(x_1, y_1)$ on a curve are equal,then what is the length of the tangent?

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