Let y=mx+c be a chord of the ellipse(x^2/a^2)+(y^2/b^2)=1
a) If the chord subtends a right angle at the point (a,0), show that (a^2+b^2)c^2+2a^3mc+a^2(a^2-b^2)m^2=0
b)Hence show that the chord passes through a fixed point on the x-axis
- 魏王將張遼Lv 71 十年前最愛解答
(a) With reference to the diagram below:
(b) The x-intercept of the chord is given by substituting y = 0 in its equation:
0 = mx + c
x = -c/m
Therefore, it can be found from the result of (a) as follows:
Since a and b are fixed, the x-intercept is also fixed and hence the result comes.資料來源： My Maths knowledge
- wyLv 71 十年前
(x^2/a^2) + (y^2/b^2) = 1.......(1)
y = mx +c ......................(2)
Substitute y of (2) into (1), we get a quadratic equation in x. Let the roots of the equation by h and k. It can be found that :
hk = a^2(c^2-b^2)/(b^2 + a^2m^2)...........(3) and
h + k = -2mca^2/(b^2 + a^2m^2).........(4)
Also, h and k are the x-coordinates of the intersecting points of the chord with the ellipse.
Since these 2 points make a right angle with (a,0), then
slope of point h with (a,0) x slope of point k with (a,0) = -1. That is
[y-coordinates of h/(h-a)][y-coordinates of k/(k-a)] = -1
y- coordinate of h x y-coordinate of k = -(h-a)(k-a)
(mh +c)(mk+c) = -(h-a)(k-a)
m^2hk + c^2 + mc(h + k) = -hk - a^2 + a(h+k)
(1+ m^2)(hk) + c^2 + a^2 + (mc - a)(h + k) = 0
Substitute the value of (3) and (4) into this equation and after simplification work, you will come to the result.
Since y= mx+ c
For y = 0, x = -c/m......(5)
Using the result of (a) part and dividing all terms by m^2, we get
(a^2 + b^2)(c/m)^2 + 2a^3(c/m) + a^2(a^2 - b^2) = 0 which is a quadratic equation in c/m.
Since a and b are constants, the roots of c/m are also constants, so x in (5) is a constant. That means the chord passes through a fixed point on the x-axis.