calculus

OK, I was just about to go and water the hanging baskets but if you insist, I shall go and get my slide rule and abacus.

I'm going to assume the test is over and proceed.

ANS: 2/√7

Just for kicks, here's how to do the problem without Calculus:

(1) Flip the ellipse over the line y = x, by substituting x for y and y for x in the equation and see what you get:

You get 2y^2-3yx+2x^2-2=0 ... so, the same ellipse as you started with. That means the ellipse is symmetric over the line y=x.

(2) Flip the ellipse over the line y = -x, by substituting -x for y and -y for x in the equation and see what you get:

You get 2(-y)^2-3(-y)(-x)+2(-x)^2-2=0 ... so, the same ellipse as you started with. That means the ellipse is symmetric over the line y = -x.

(3) These results mean: (a) the centre of the ellipse is (0,0); and (b) the two axes overlap the lines y=x and y= -x. The point(s) at a minimum distance from (0,0) must lie on one of these lines.

(4) Find the points of intersection between y=x and the ellipse by subbing in x for y in the ellipse's equation. You get 2x^2 - 3x^2 + 2x^2 = 2, or x^2 = 2. Which means (x,y) = (±√2, ±√2). The distance from either point to (0,0) is 2.

(5) Find the point of intersection between y = -x and the ellipse by subbing in -x for y in the ellipse equation. You get 2x^2 + 3x^2 + 2x^2 = 2, or 7x^2 = 2. Which means (x,y) = (±√(2/7), ∓√(2/7)). The distance from either point to (0,0) is 2/√7. That's the minimum distance.