Question

The gradient of the curve y-xy+2px+3qy=0 at (1,1) is 1/2.The values of p and q are​

Answer 1

Answer:

please give me alok please

Answer 2

Answer:

at first the point (3,2) will satisfy this curve equation so,

(2) -(3)(2)+2p(3)+3q(2) = 0

=> 2-6+6p+6q = 0

=> (p+q) = 4/6 = 2/3

also, gradient is = dy/dx = -2/3

dy/dx - (y+xdy/dx) + 2p+3qdy/dx = 0

=> (-2/3)-(2+3*(-2/3)) +2p +3q(-2/3) = 0

=> 2p -2q = 2/3

=> p-q = 1/3

so, adding both equation.

=> 2p = 1. or p = 1/2 , and q = 2/3-1/2 = 1/6

so p = 1/2....q = 1/6