Question

The equation of the curve $y=f\left(x\right)$ in first quadrant with positive tangent slope such that the sum of the lengths of the tangent and subtangent at any point on it is proportional to the product of the coordinates of the point (proportionality factor is $k$) is:
(where $c$ is arbitrary constant)

• A. $y=\frac{1}{k}\ln \left|c\right(k^2{x}^2-1\left)\right|$
• B. $y=\frac{1}{k}\ln \left|c\right(k{x}^2-1\left)\right|$
• C. $y=\frac{1}{k}\ln \left|c\right(k^2{x}^2+1\left)\right|$
• D. $y=\ln \left|c\right(k^2{x}^2+1\left)\right|$

Answer 1

The correct option is A $y=\frac{1}{k}\ln \left|c\right(k^2{x}^2-1\left)\right|$

Length of tangent at any point of curve $=y\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}$
Length of subtangent $=y\frac{dx}{dy}$
Given: $y\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}+y\frac{dx}{dy}=kxy$
$\Rightarrow \sqrt{1+{\left(\frac{dx}{dy}\right)}^2}=kx-\frac{dx}{dy}$
$\Rightarrow 1+{\left(\frac{dx}{dy}\right)}^2=k^2{x}^2+{\left(\frac{dx}{dy}\right)}^2-2kx\frac{dx}{dy}$
$\Rightarrow 2kx\frac{dx}{dy}=k^2{x}^2-1\Rightarrow \int \frac{2kxdx}{k^2{x}^2-1}=\int dy$
On integrating we get:
$y=\frac{1}{k}\ln \left|c\right(k^2{x}^2-1\left)\right|$