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

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

A

(y−xdydx)2=1−(dydx)2\left( y - x \frac { d y } { d x } \right) ^ { 2 } = 1 - \left( \frac { d y } { d x } \right) ^ { 2 }

B

(y+xdydx)2=1+(dydx)2\left( y + x \frac { d y } { d x } \right) ^ { 2 } = 1 + \left( \frac { d y } { d x } \right) ^ { 2 }

C

(y−xdydx)2=1+(dydx)2\left( y - x \frac { d y } { d x } \right) ^ { 2 } = 1 + \left( \frac { d y } { d x } \right) ^ { 2 }

D

(y+xdydx)2=1−(dydx)2\left( y + x \frac { d y } { d x } \right) ^ { 2 } = 1 - \left( \frac { d y } { d x } \right) ^ { 2 }

Answer

(y−xdydx)2=1+(dydx)2\left( y - x \frac { d y } { d x } \right) ^ { 2 } = 1 + \left( \frac { d y } { d x } \right) ^ { 2 }

Explanation

Solution

Since the equation of lines whose distance from origin is unit, is given by xcos⁡α+ysin⁡α=1x \cos \alpha + y \sin \alpha = 1 .....(i)

Differentiate w.r.t. x, we get cos⁡α+dydxsin⁡α=0\cos \alpha + \frac { d y } { d x } \sin \alpha = 0 .....(ii)

On eliminating the with the help of (i) and (ii)

i.e., (i) –x × (ii)

⇒ sin⁡α(y−xdydx)=1\sin \alpha \left( y - x \frac { d y } { d x } \right) = 1 ⇒ (y−xdydx)=cosec⁡α\left( y - x \frac { d y } { d x } \right) = \operatorname { cosec } \alpha .....(iii)

Also (ii) ⇒ dydx=−cot⁡α\frac { d y } { d x } = - \cot \alpha ⇒ (dydx)2=cot⁡2α\left( \frac { d y } { d x } \right) ^ { 2 } = \cot ^ { 2 } \alpha .....(iv)

Therefore by (iii) and (iv), 1+(dydx)2=(y−xdydx)21 + \left( \frac { d y } { d x } \right) ^ { 2 } = \left( y - x \frac { d y } { d x } \right) ^ { 2 }.