The differential equation for all the straigh | vedclass

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

Question

(C) The equation of a straight line at a unit distance from the origin is given by $x \cos \alpha + y \sin \alpha = 1$ $... (i)$Differentiating with r

Vedclass · Accepted Answer

${\left( {y - x\frac{{dy}}{{dx}}} \right)^2} = 1 + {\left( {\frac{{dy}}{{dx}}} \right)^2}$

Vedclass · Answer

(C) The equation of a straight line at a unit distance from the origin is given by $x \cos \alpha + y \sin \alpha = 1$ $... (i)$Differentiating with respect to $x$,we get $\cos \alpha + \frac{dy}{dx} \sin \alpha = 0$ $... (ii)$From $(ii)$,we have $\cos \alpha = -\frac{dy}{dx} \sin \alpha$. Substituting this into $(i)$:$x(-\frac{dy}{dx} \sin \alpha) + y \sin \alpha = 1$$\sin \alpha (y - x \frac{dy}{dx}) = 1$$y - x \frac{dy}{dx} = \frac{1}{\sin \alpha} = \csc \alpha$ $... (iii)$From $(ii)$,$\frac{dy}{dx} = -\cot \alpha$,so $(\frac{dy}{dx})^2 = \cot^2 \alpha$.Using the identity $1 + \cot^2 \alpha = \csc^2 \alpha$,we substitute from $(iii)$ and $(iv)$:$1 + (\frac{dy}{dx})^2 = \csc^2 \alpha = (y - x \frac{dy}{dx})^2$.

The differential equation for all the straight lines which are at a unit distance from the origin is

Similar Questions

The order of the differential equation whose solution is $y=a \cos x+b \sin x+c e^{-x}$ is

The differential equation of all straight lines passing through the point $(1, -1)$ is

The differential equation of the family of parabolas with vertex at $(0,-1)$ and having axis along the $Y$-axis is

The differential equation of all parabolas whose axis is the $y$-axis is:

Let $p \in \mathbb{R}$. Then the differential equation of the family of curves $y=(\alpha+\beta x) e^{p x}$, where $\alpha$ and $\beta$ are arbitrary constants, is

Vedclass Test Series

Exam Paper Generator

Online Exam Module