If `ax^(2)+2hxy+by^(2)+2gx+2fy+c=0`, then show that `(dy)/(dx).(dx)/(dy) = 1`.

Answer

N/a

Explanation

We have, `ax^(2)+2hxy+by^(2)+2gx+2fy+c=0`
On differentiating both sides w.r.t.x, we get
`(d)/(dx)(ax)^(2)+(d)/(dx)(2hxy)+(d)/(dx)(by^(2))+(d)/(dx)(2gx)+(d)/(dx)(2fy)+(d)/(dx)(c) = 0`
`rArr (dy)/(dx)[2hx+2by+2f]= - 2ax-2hy-2g`
`rArr (dy)/(dx) = (-2(ax+hy+g))/(2(hx+by+f))`
`= (-(ax+hy+g))/((hx+by+f))"......"(ii)`
Now, differentiating Eq. (i) w.r.t y., we get
`(d)/(dy)(ax^(2))+(d)/(dy)(2hxy)+(d)/(dy)(by^(2))+(d)/(dy)(2gx)+(d)/(dy)(2fy)+(d)/(dy)(c)= 0`
`rArr a.2x.(dx)/(dy)+2h.(x.(d)/(dy)y+y.(d)/(dy)x)+b.2y+2g.(dx)/(dy)+2f+0=0`
`rArr (dx)/(dy)[2ax+2hy+2g]=-2hx-2by-2f`
`rArr (dx)/(dy) = (-2(hx+by+f))/(2(ax+hy+g)) = (-(hx+by+f))/((ax+hy+g))"......"(ii)`
`:. (dy)/(dx).(dx)/(dy)= (-(ax+hy+g))/((hx+by+f)).(-(hx+by+f))/((ax+hy+g))` [using Eqs. (ii) and (iii)]
` = 1 = RHS