It is known that the straight line l passes through the parabola y? =2px focus F( 1, 0), parabola intersecting at point M and point N, O as the origin of coordinates, and lines MO and NO intersecting at point P and point Q respectively to find the minimum value of PQ.

Solution: ∫c = p/2 = 1, ∴p=2, so the parabolic equation is y? =4x .

Let the linear equation of the focus F be y=k(x- 1), and substitute it into the parabolic equation to get k? (x- 1)？ =4x, which means there is a k? x？ -2(k？ +2)x+k？ =0;

Let M(x? ，y？ )，N(x？ ，y？ ), and then x? +x？ =2(k？ +2)/k？ ; x？ x？ = 1;

y？ +y？ =k(x？ +x？ )-2=2(k？ +2)/k-2=(2k？ -2k+4)/k=2(k？ -k+2)/k。

y？ y？ =k？ (x？ - 1)(x？ - 1)=k？ 【x？ x？ -(x？ +x？ )+ 1]=k？ 【 1-2(k？ +2)/k？ + 1]=-4。

x？ = y/4； x？ = y/4；

The equation of the straight line where MO is located is y=(y? /x？ )x=(4/y？ )x； The equation of the line where NO is located is y=(y? /x？ )x=(4/y？ )x；

Because the directrix equation is x=- 1, let x=- 1, and the coordinates of the intersection point p between Mo straight line and directrix are (-1, -4/y? ); The coordinates of Q at the intersection of the line where NO is located and the directrix are (-1, -4/y? );

What about ∣PQ∣=∣-4/y? +4/y？ ∣=4∣(y？ -Really? )/y？ y？ ∣=4∣[(y？ +y？ )? -4y？ y？ ]/(y？ y？ )∣

=4∣{[4(k？ -k+2)？ /k？ ]+ 16}/(-4)∣=∣[4(k？ -k+2)？ /k？ ]+ 16∣=4∣[(k？ -k+2)？ /k？ ]+4∣

=4∣(k- 1+2/k)？ +4∣=4∣[k+(2/k)- 1]？ +4∣≧4∣(2√2- 1)? +4∣=4(8-4√2+ 1+4)=4( 13-4√2)

When k=2/k, k? =2 and k = √ 2. That is, when k = √ 2, PQ gets the minimum value of 4( 13-4√2).