37. It is known that the parabola $C : x ^ { 2 } = 4 y$, the line through the point $M ( 0,4 )$ and $C$ intersects at the points $A , B$, and the tangent line at the points $C _ { \text {在 } } A , B$ intersects at the point $N$. Of the following four points, the one that can be the midpoint of the line segment $M N$ is ( )
- A. A. $( 0,1 )$
- B. B. $\left( \frac { 1 } { 2 } , 0 \right)$
- C. C. $\left( \frac { 1 } { 2 } , 1 \right)$
- D. D. $\left( 1 , - \frac { 1 } { 2 } \right)$
Answer: B
Solution: It may be useful to set $A \left( x _ { 1 } , y _ { 1 } \right) , B \left( x _ { 2 } , y _ { 2 } \right) , N \left( x _ { 0 } , y _ { 0 } \right)$, and from $x ^ { 2 } = 4 y$, $y = \frac { 1 } { 4 } x ^ { 2 }$, then $y ^ { \prime } = \frac { 1 } { 2 } x$.
The equation of the tangent to the parabola at the point $A$ is then $y - y _ { 1 } = \frac { 1 } { 2 } x _ { 1 } \left( x - x _ { 1 } \right)$, and
Since $x _ { 1 } ^ { 2 } = 4 y _ { 1 }$, the equation is simplified to $x _ { 1 } x = 2 \left( y + y _ { 1 } \right)$ and
Similarly, the equation of the tangent to the parabola at the point ${ } ^ { B }$ is $x _ { 2 } x = 2 \left( y + y _ { 2 } \right)$, and
And the tangent lines intersect at the point $N \left( x _ { 0 } , y _ { 0 } \right)$, so we get $\left\{ \begin{array} { l } x _ { 1 } x _ { 0 } = 2 \left( y _ { 0 } + y _ { 1 } \right) \\ x _ { 2 } x _ { 0 } = 2 \left( y _ { 0 } + y _ { 2 } \right) \end{array} \right.$.
i.e., the points $A , B$ are on the line $x _ { 0 } x - 2 y - 2 y _ { 0 } = 0$, i.e., the equation of the line $A B$ is $x _ { 0 } x - 2 y - 2 y _ { 0 } = 0$, and $x _ { 0 } x - 2 y - 2 y _ { 0 } = 0$ is $x _ { 0 } x - 2 y - 2 y _ { 0 } = 0$.
Since the point $M ( 0,4 )$ is on the line $A B$, substituting solves for $y _ { 0 } = - 4$, which is $N \left( x _ { 0 } , - 4 \right)$.
Therefore, the midpoint of segment $M N$ is $N \left( \frac { x _ { 0 } } { 2 } , 0 \right)$, so the option that can be the midpoint of segment $M N$ is $\left( \frac { 1 } { 2 } , 0 \right)$.
