For the curve 4x^{5} = 5y^{4},the ratio of th | vedclass

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

Question

(C) Given curve: $4x^{5} = 5y^{4}$.Differentiating with respect to $x$: $20x^{4} = 20y^{3} \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{x^{4}}{y^{3}}$

Vedclass · Accepted Answer

$\frac{4^{4}}{5^{4}}$

Vedclass · Answer

(C) Given curve: $4x^{5} = 5y^{4}$.Differentiating with respect to $x$: $20x^{4} = 20y^{3} \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{x^{4}}{y^{3}}$.Length of subtangent $(ST) = \left| y \frac{dx}{dy} \right| = \left| y \frac{y^{3}}{x^{4}} \right| = \frac{y^{4}}{x^{4}}$.Length of subnormal $(SN) = \left| y \frac{dy}{dx} \right| = \left| y \frac{x^{4}}{y^{3}} \right| = \frac{x^{4}}{y^{2}}$.We need the ratio of the cube of the subtangent to the square of the subnormal:$\frac{(ST)^{3}}{(SN)^{2}} = \frac{(y^{4}/x^{4})^{3}}{(x^{4}/y^{2})^{2}} = \frac{y^{12}/x^{12}}{x^{8}/y^{4}} = \frac{y^{16}}{x^{20}} = \left(\frac{y^{4}}{x^{5}}\right)^{4}$.Since $4x^{5} = 5y^{4}$,we have $\frac{y^{4}}{x^{5}} = \frac{4}{5}$.Therefore,the ratio is $\left(\frac{4}{5}\right)^{4} = \frac{4^{4}}{5^{4}}$.

For the curve $4x^{5} = 5y^{4}$,the ratio of the cube of the subtangent at a point on the curve to the square of the subnormal at the same point is

Similar Questions

If the equation of the tangent at $(2,3)$ on the curve $y^2 = ax^3 + b$ is $y = 4x - 5$,then the value of $a^2 + b^2$ is:

The ratio between the length of the subtangent at any point other than the origin on the parabola $y^2 = 16ax$ and the abscissa of that point is:

The equation of the tangent to the curve $y = \cos(x + y)$ for $-2\pi \leq x \leq 2\pi$,which is parallel to the line $x + 2y = 0$,is:

If $\theta$ is the angle between the curves $xy=2$ and $x^2+4y=0$,then $\tan \theta$ is equal to:

The equation of the normal to the curve $y = \log_e x$ at the point $P(1, 0)$ is $ . . . . . . $

Vedclass Test Series

Exam Paper Generator

Online Exam Module