Differential Equations Question 344
Question: The equation of family of curves for which the length of the normal is equal to the radius vector is
Options:
A) $ y^{2}\pm x^{2}=k $
B) $ y\pm x=k $
C) $ y^{2}=kx $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Length of the normal $ =y\sqrt{1+{{( \frac{dy}{dx} )}^{2}}} $
It is given that $ y\sqrt{1+{{( \frac{dy}{dx} )}^{2}}}=\sqrt{x^{2}+y^{2}} $ ( $ \because $ Radius vector $ =r=\sqrt{x^{2}+y^{2}} $ )
Therefore $ y^{2}+y^{2}{{( \frac{dy}{dx} )}^{2}}=x^{2}+y^{2} $
Therefore $ y^{2}{{( \frac{dy}{dx} )}^{2}}=x^{2} $
Therefore $ ydy\pm xdx=0 $
Therefore $ y^{2}\pm x^{2}=k $ .
BETA
