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 :
[c] \[y=mx+c\] (Equation of straight lie) \[\frac{dy}{dx}=m\] and \[mx-y+c=0\] is at unit distance from origin. \[\therefore \frac{\left| m(0)-(0)+c \right|}{\sqrt{{{m}^{2}}+{{(-1)}^{2}}}}=1\Rightarrow c=\sqrt{1+{{m}^{2}}}\] Now: \[{{\left[ y-x\frac{dy}{dx} \right]}^{2}}={{[mx+c-xm]}^{2}}={{c}^{2}}=1+{{m}^{2}}\] also, \[{{\left[ y+x\frac{dy}{dx} \right]}^{2}}={{[mx+c\,+mx]}^{2}}={{[2mx+\sqrt{1+{{m}^{2}}}]}^{2}}\] also, \[1-{{\left( \frac{dy}{dx} \right)}^{2}}=1-{{m}^{2}}\] and \[1+{{\left( \frac{dy}{dx} \right)}^{2}}=1+{{m}^{2}}\] \[\Rightarrow {{\left[ y-x\frac{dy}{dx} \right]}^{2}}=1+{{\left( \frac{dy}{dx} \right)}^{2}}\]
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