Let two points A (X 1, Y 1) and B (X2, Y2) exist.

Then the midpoint M(x0, y0) of AB is within the ellipse and on the straight line y = 4x+m..

AB is perpendicular to the straight line y=4x+m

List known relationships:

3x12+4y12 =12 ...1(A is on the ellipse)

3x2x2+4Y22 =12 ... 2 (B is on the ellipse)

2x0 = x 1+x2...3 (m is the midpoint of AB)

2 y0 = y 1+y2...4 (same as above)

Y0 = 4x0+m...5 (m on a straight line, y=4x+m)

(y1-y2)/(x1-x2) =-1/4 ... 6 (ab is perpendicular to the straight line y=4x+m)

3x0^2+4y0^2<； 12...7(M in ellipse)

1 formula -2 formula:

3(x 1-x2)(x 1+x2)+4(y 1-y2)(y 1+y2)= 0

Using equations 3, 4 and 6, we can get

3x0-y0=0

Combining Equation 5, you can get:

x0=-m，y0=-3m

Substitute into equation 7:

3m^2+4(3m)^2=39m^2<； 12

So m 2

So-2 √ 13/ 13

2.

The curve x 2+y 2+2x-6y+ 1 = 0 is a circle.

So PQ is a string.

The straight line in the problem is the middle vertical line of the chord,

Therefore, it is the straight line where the diameter of the circle lies.

Pass through the center M(- 1, 3). Substitution can get m=- 1.

The equation of the straight line x+my+4=0 is x-y+4=0.

And PQ is perpendicular to the straight line x-y+4=0,

Let its equation be x+y+b=0.

Let P(X 1, Y 1) and Q(x2, y2),

Because OP⊥OQ, there is x 1x2+y 1y2=0.

Put y=-x-b

2x 2+(8+2b) x+b 2+6b+ 1 = 0，

have

x 1x2=(b^2+6b+ 1)/2,x 1+x2=-(4+b)，

y 1y2=x 1x2+b(x 1+x2)+b^2=(b^2+6b+ 1)/2-b(4+b)+b^2=(b^2-2b+ 1)/2。

And (B2+6b+1)/2+(B2-2b+1)/2 = 0.

B 2+2 b+ 1 = 0, that is, b=- 1.

So the PQ equation is x+y- 1=0.