Poly Mathsssss

• Wireless-

  13 Feb 10, 16:03

  

• jayh272416

  13 Feb 10, 16:15

  discriminant = 0

  take it as a quadratic equation in terms of y. find the max x value such that the equation can be solved.

• wee_ws

  13 Feb 10, 16:18

  Hi,

    1. Differentiate the equation implicitly with respect to x.

    2. Express dy/dx in terms of x and y, and obtain an expression with numerator and denominator.

    3. Equate the denominator of dy/dx to zero (why?).

    4. Substitute back into the equation, to find x, which will be less than 3.

    5. The required distance is their difference.

    Give it a try, jiayou!

  Cheers,
  Wen Shih

• eagle

  13 Feb 10, 16:25

  Hi, if you have topics next time, please include it as well becaues we might use equations which you haven't learned.

   

  The point nearest to the wall should be dependent on x in this case.

  So simply, we can make x the subject, and minus 3 from it. You will get an equation in y which you need to find the minimum. So you can use differentiation to solve it

  But... this method is too tedious... so to 'play cheat', I would use a different method....
  The method is.... find the points (x,y) such that the gradient is vertical. Just imagine you moved the vertical wall to touch the curve.

  Ans:

  Differentiate with respect to x
  2x - y - x dy/dx + 2y dy/dx = 0
  dy/dx (2y - x) = y - 2x
  dy/dx = (y - 2x) / 2y - x
  so for dy/dx to be infinity for vertical gradient, 2y = x
  
  Sub x = 2y into x^2 - xy + y^2 = 3
  So 4 y^2 - 2y^2 + y^2 = 3
  3 y^2 = 3
  y = 1 or -1

  Hence, x = 2 or -2

  Choose the one with x as positive (see from graph)

  Hence, (2, 1) is the point nearest to the wall at x = 3, and the nearest distance is 1 unit.

• Wireless-

  13 Feb 10, 16:50

  omg thanks so much for quick reply. (2,1) is correcT. understood!

• yiha093

  16 Feb 10, 03:47

  ya la... see girl then all reply so fast..... xD