AB is a focal chord of the parabola y^2 = 4x, and the coordinates of A are (4,4). Through A and B, normals are drawn meeting the parabola again at A’ and B’, respectively. Prove that A’B’ is three times as long as AB.

For the horizontal parabola y²=4x=4px where p is the distance of the vertex (0,0) (in this case) from the focus. So p=1 and the focus is at (1,0). We now have two points ((1,0) and A(4,4)) on the focal chord so we can work out the slope and the equation. Slope=(4-0)/(4-1)=4/3. The equation of the chord is y=(4/3)(x-1), y=4x/3-4/3.

Now we can work out where B is.

Substitute x=y²/4: y=y²/3-4/3, y²-3y-4=(y-4)(y+1). Obviously A is one point so B must be on the chord line where y=-1, so x=y²/4=1/4, making B(1/4,-1).

Since x=y²/4, 1=(y/2)dy/dx, dy/dx=2/y, so at A the tangent is 2/4=1/2 and at B it is -2. We can work out the slope of the normals: -2 and 1/2 respectively. The equations of the normals are at A: y-4=-2(x-4), y=-2x+12; at B: y+1=(1/2)(x-1/4), y=x/2-9/8.

Now we need to find A' and B'.

Substitute x=y²/4 in the normal equation for A: y=-y²/2+12, y²+2y-24=0=(y-4)(y+6). We already know y=4 is a solution because of A, so the normal intersects the parabola at y=-6 and x=36/4=9. So A'(9,-6).

Substitute x=y²/4 in the normal equation for B: y=y²/8-9/8, y²-8y-9=(y+1)(y-9). We already know y=-1 is a solution because of B, so the normal intersects the parabola at y=9 and x=81/4. So B'(81/4,9).

Length of AB²=(4-1/4)²+(4-(-1))²=625/16, AB=25/4.

Length of A'B'²=(81/4-9)²+(9-(-6))²=5625/16, A'B'=75/4 which 3AB (QED).

answered May 14 by Top Rated User (559,640 points)