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.
