If the length of the subtangent and the lengt | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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?

Similar Questions

What is the slope of the tangent to the hyperbola $2x^2 - 3y^2 = 6$ at the point $(3, 2)$?

If the tangent to the curve $y = \frac{x}{x^2-3}$,$x \in R, (x \neq \pm \sqrt{3})$ at a point $(\alpha, \beta) \neq (0,0)$ on it,is parallel to the line $2x + 6y - 11 = 0$,then

The equation of the tangent to the curve $y=\sqrt{9-2x^2}$ at the point where the ordinate and abscissa are equal is:

The angle between the curves $y^2=4x+4$ and $y^2=36(9-x)$ is (in $^{\circ}$)

Find the points on the curve $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$ at which the tangents are parallel to the $y$-axis.

Vedclass Test Series

Exam Paper Generator

Online Exam Module