PE+PF=2a

PE+PQ=EQ=2a

So PF=PQ

That is, △PFQ is an isosceles triangle.

The product of the number of vectors PT and TF is equal to 0.

Namely PT⊥TF

So TF=TQ

That is, t is the midpoint of QF.

Let P(x 1, y 1) and T(x, y).

Because |EQ|=2a

That is (x 1+c)? +? (y 1)？ =4a？

T is the midpoint of QF.

So x 1+c=2x.

y 1=2y

Bring in the above formula

simplify

x？ +? y？ =a？

So the trajectory of point T is a circle with the origin as the center and a as the radius.

Let m coordinate be (m, n)

Then the area of △EMF is s = 1/2ef * | n | = b 2.

That is c | n | = b 2

|n|=b^2/c

When b 2/c ≤ a

That is, when a≤( 1-√5)c/2.

There is such a point m.

At this time, due to the symmetry of the ellipse, there should be two or four such points.

Let's take m in the first quadrant or on the positive semi-axis of the y axis as an example.

At this time, M([ under the radical sign (A 2C 2-B 4)]/c, B 2/c)

Then use the included angle formula of the straight line to find out.

When b 2/c > A

That is, a>( 1-√5)c/2.

There is no such point m.