Which of the following pair (s) `is//are` orthogonal ?

Options

A. `16x^(2)+y^(2)=c and y^(16)=kx`
B. `y=x+ce^(-x) and x+2=y+ke^(-y)`
C. `y=cx^(2) and x^(2)+2y^(2)=k`
D. `x^(2)-y^(2)=c and xy =k`

Answer

A, B, C, D

Explanation

(a) `16x^(2)+y^(2)=c rArr(dy)/(dx)=m_(1)=-(16x)/(y)`
`y^(16)=kx" "rArr " "(dy)/(dx)=m_(2)=(k)/(16y^(15))`
`m_(1)m_(2)=-(16x)/(y).(k)/(16y^(15))`
`=-(x)/(y^(16)).k`
`=-(x)/(y^(16)).(y^(16))/(x)=-1`
`rArr" Curves are orthogonal."`
(b) `y=x+ce^(-x)`
`rArr" "(dy)/(dx)=1-ce^(-x)=1-(y-x)=-(y-x-1)`
`x+2=y+ke^(-y)`
`rArr" "(dy)/(dx)-k(dy)/(dx)e^(-y)=1`
`rArr" "(dy)/(dx)[1-ke^(-y)]=1`
`"or "[1-(x+2-y)](dy)/(dx)=1`
`rArr" "(dy)/(dx)=m_(2)=(1)/(y-x-1)`
`rArr: "m_(1)m_(2)=-1`
`rArr" Curves are orthogonal."`
(c) `y=cx^(2)`
`rArr" "(dy)/(dx)=2cx=2x(y)/(x^(2))=(2y)/(x)=m_(1)`
`"and "x^(2)+2y^(2)=k`
`rArr" "2x+4y(dy)/(dx)=0`
`rArr" "(dy)/(dx)=-(x)/(2y)=m_(2)`
`rArr" "m_(1)m_(2)=-1`
`rArr" Curves are orthogonal."`
(c) `x^(2)-y^(2)=c rArr (dy)/(dx)=(x)/(y)=m_(1)`
`xy=k rArr (dy)/(dx)=-(y)/(x)=m_(2)`
`therefore" "m_(1)m_(2)=-1`
`rArr" Curves are orthogonal."`