The correct option is A y=cx−tan−1√1+c2
√1+p2=tan(px−y) ...(1)
put x=0, we get √1+p2=−tany(0)
⇒y(0)=−tan−1√1+p2 ....(2)
now, squaring (1) & then differentiating, we get
2(1+p2)pdpdx=2tan(px−y)sec2(px−y)(p+xdpdx−dydx)
⇒2(1+p2)pdpdx=2tan(px−y)sec2(px−y)xdpdx
Therefore, dpdx=0 satisfies the above equation.
by integrating p=c
⇒dydx=c
by integrating y=cx+b
put x=0, we get y(0)=b
from (2): b=−tan−1√1+p2=−tan−1√1+c2
Therefore, y=cx−tan−1√1+c2
Ans: A