Differential Equations Question 104
Question: The differential equation for all the straight lines which are at a unit distance from the origin is
[MP PET 1993]
Options:
A) $ {{( y-x\frac{dy}{dx} )}^{2}}=1-{{( \frac{dy}{dx} )}^{2}} $
B) $ {{( y+x\frac{dy}{dx} )}^{2}}=1+{{( \frac{dy}{dx} )}^{2}} $
C) $ {{( y-x\frac{dy}{dx} )}^{2}}=1+{{( \frac{dy}{dx} )}^{2}} $
D) $ {{( y+x\frac{dy}{dx} )}^{2}}=1-{{( \frac{dy}{dx} )}^{2}} $
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Answer:
Correct Answer: C
Solution:
Since the equation of lines whose distance from origin is unit, is given by $ x\cos \alpha +y\sin \alpha =1 $
…..(i) Differentiate w.r.t. x, we get $ \cos \alpha +\frac{dy}{dx}\sin \alpha =0 $ …..(ii) On eliminating the $ ‘\alpha ’ $ with the help of (i) and (ii) i.e., (i) -x × (ii)
Therefore $ \sin \alpha ( y-x\frac{dy}{dx} )=1 $
Therefore $ ( y-x\frac{dy}{dx} )=cosec\alpha $ …..(iii) Also (ii)
Therefore $ \frac{dy}{dx}=-\cot \alpha $
Therefore $ {{( \frac{dy}{dx} )}^{2}}={{\cot }^{2}}\alpha $
…..(iv) Therefore by (iii) and (iv), $ 1+{{( \frac{dy}{dx} )}^{2}}={{( y-x\frac{dy}{dx} )}^{2}} $ .
BETA
