/br>
如图,由抛物线的定义,得|MF|=|MM 1 |,|NF|=|NN 1 |.
∴∠MFM 1 =∠MM 1 F,∠NFN 1 =∠NN 1 F.
设准线l与x轴的交点为F 1 ,∵MM 1 ∥ FF 1 ∥ NN 1 ,
∴∠MM 1 F=∠M 1 FF 1 ,∠NN 1 F=∠N 1 FF 1 .
而∠MFM 1 +∠M 1 FF 1 +∠NFN 1 +∠N 1 FF 1 =180°,
∴2∠M 1 FF 1 +2∠N 1 FF 1 =180°,即∠M 1 FN 1 =90°.
答案 C