Ellipse - Ans.fyi 參考答案

Ellipse

用戶182392

2007-01-29 6:59 am

The ellipse E has equation x^2/a^2+y^2/b^2=1 and the line L has equation =mx+c, where m>0 and c>0. If L is a tangent to the ellipse and meets the negative x-axis at the point A and the positive y-axis at the point B, and O is the origin,
(i) Prove that, as m varies, the minimum area of triangle OAB is ab.
(ii) Find, in terms of a, the x-coordinate of the point of contact of L and E when the area of triangle OAB is a minimum.

回答 (1)

用戶127502

2007-01-29 10:35 am

✔ 最佳答案

(i)
Let P is the Point of contact of L and E = (x0,y0)
Then L: (x0/a^2) x + (y0/b^2) y = 1
Rearrange, as L: y=mx+c
m = - (b^2/a^2)(x0/y0), c = b^2/y0 -----(*)

Area = (1/2) (c) (c/m)
Sub (*),
Area = (-1/2) (a^2b^2) / (x0y0)
Consider (x0/a + y0/b)^2 >=0
x0^2/a^2 + 2(x0y0)/(ab) + y0^2/b^2 >=0
1+2(x0y0)/(ab) >=0
x0y0 >= (-1/2) ab
Therefore
Area = (-1/2) (a^2b^2) / (x0y0) >= ab
and the result follows.

(ii) At minimum, equality holds.
x0y0 = (-1/2) ab
With the ellipse E, by simultaneous equations in two unknown
x0 = a/sqrt2

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