Question

Assertion :The differential equation of all straight lines which are at a constant distance $p$ from the origin is ${(y-x{y}_1)}^2=p^2(1+y_1^2)$ Reason: The general equation of any straight line which is at a constant distance $p$ from the origin is $x\cos \alpha +y\sin \alpha =p.$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Any straight line which is at a constant distance $p$ from the origin is

$x\cos \alpha +y\sin \alpha =p$ ...(1)

Differentiating both sides w.r.t ${}^{\prime }{x}^{\prime },$ we get

$\cos \alpha +\sin \alpha \frac{dy}{dx}=0$

$\Rightarrow \tan \alpha =\frac{-1}{y_1}$ where $y_1=\frac{dy}{dx}$

$\therefore \sin \alpha =\frac{2}{\sqrt{1+y_1^2}};\cos \alpha =\frac{-y_1}{\sqrt{1+y_1^2}}$

Putting in (1), we get

$x.\frac{-y_1}{\sqrt{1+y_1^2}}+y\frac{1}{\sqrt{1+y_1^2}}=p$

$\Rightarrow {(y-{xy}_1)}^2=p^2(1+y_1^2)$