The length of the subtangent at any point of | vedclass

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

Question

(B) Given the curve equation: $x^m y^n = a^{m+n}$.Taking the natural logarithm on both sides: $m \ln x + n \ln y = \ln(a^{m+n})$.Differentiating with

Vedclass · Accepted Answer

Abscissa

Vedclass · Answer

(B) Given the curve equation: $x^m y^n = a^{m+n}$.Taking the natural logarithm on both sides: $m \ln x + n \ln y = \ln(a^{m+n})$.Differentiating with respect to $x$: $\frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = 0$.Thus,$\frac{dy}{dx} = -\frac{m}{n} \frac{y}{x}$.The length of the subtangent is given by the formula: $LST = \left| \frac{y}{dy/dx} \right|$.Substituting the value of $\frac{dy}{dx}$: $LST = \left| \frac{y}{(-\frac{m}{n} \frac{y}{x})} \right| = \left| -\frac{nx}{m} \right| = \frac{n}{m} |x|$.Since $\frac{n}{m}$ is a constant,the length of the subtangent is proportional to the abscissa $x$.

The length of the subtangent at any point of the curve $x^my^n = a^{m+n}$ is proportional to

Similar Questions

Find the equation of all lines having slope $2$ which are tangents to the curve $y = \frac{1}{x-3}, x \neq 3$.

If the curves $x^2+p y^2=1$ and $q x^2+y^2=1$ are orthogonal to each other,then

An angle between the curves $x^2-y^2=4$ and $x^2+y^2=4 \sqrt{2}$ is

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

If $ab \neq 0$,then the equation of the tangent at $(a, b)$ to the curve $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module