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.
asked May 13 in Calculus Answers by anonymous

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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 Rod Top Rated User (537,180 points)

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