A) \[{{\left( y-x\frac{dy}{dx} \right)}^{2}}=1-{{\left( \frac{dy}{dx} \right)}^{2}}\]
B) \[{{\left( y+x\frac{dy}{dx} \right)}^{2}}=1+{{\left( \frac{dy}{dx} \right)}^{2}}\]
C) \[{{\left( y-x\frac{dy}{dx} \right)}^{2}}=1+{{\left( \frac{dy}{dx} \right)}^{2}}\]
D) \[{{\left( y+x\frac{dy}{dx} \right)}^{2}}=1-{{\left( \frac{dy}{dx} \right)}^{2}}\]
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) Þ \[\sin \alpha \left( y-x\frac{dy}{dx} \right)=1\] Þ\[\left( y-x\frac{dy}{dx} \right)=\text{cosec}\alpha \] .....(iii) Also (ii) Þ \[\frac{dy}{dx}=-\cot \alpha \] Þ \[{{\left( \frac{dy}{dx} \right)}^{2}}={{\cot }^{2}}\alpha \] .....(iv) Therefore by (iii) and (iv), \[1+{{\left( \frac{dy}{dx} \right)}^{2}}={{\left( y-x\frac{dy}{dx} \right)}^{2}}\].
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