The normal at a general point (a, b) on curve makes an angle `theta` with x-axis which satisfies `b(a^(2)tan theta - cot theta)=a(b^(2)+1)`. The equation of curve can be

Options

A. `y = e^(x^(2)//2)+c`
B. `log ky^(2) = x^(2)`
C. `y = ke^(x^(2)//2)`
D. `x^(2)-y^(2)=k`

Answer

B, C, D

Explanation

Slope of normal `tan theta = -(dx)/(dy)`
`therefore" "`The given equation becomes at a general point (x, y)
as`" "y(x^(2)(-(dx)/(dy))+(dy)/(dx))=x(y^(2)+1)`
`rArr" "-yx^(2)+y((dy)/(dx))^(2)=(dy)/(dx)*x(y^(2)+1)`
`rArr" "y((dy)/(dx))^(2)-x(y^(2)+1)(Dy)/(dx)-yx^(2)=0`
`rArr" "yy'^(2) - xy^(2)y'-xy'-yx^(2)=0`
`rArr" "yy'(y'-xy)-x(y'-xy)=0`
`therefore" "(dy)/(dx)=xy or (dy)/(dx)=(x)/(y)`
`therefore" "log y = (x^(2))/(2)+c(or) x^(2) - y^(2) = c`
`therefore" "y = ke^((x^(2))/(2)) (or) log y^(2) = x^(2) - log k`
Also `y = k*e^((x^(2))/(2)) rArr log ky^(2) = x^(2)`