y1=-x12+
9 |
2 |
①代入②得x12+(k-
9 |
2 |
∵P為切點(diǎn),
∴△=(k-
9 |
2 |
17 |
2 |
1 |
2 |
當(dāng)k=
17 |
2 |
當(dāng)k=
1 |
2 |
∵P在第一象限,∴所求的斜率k=
1 |
2 |
(2)過P點(diǎn)作切線的垂線,其方程為y=-2x+5③
將③代入拋物線方程得x2-
13 |
2 |
設(shè)Q點(diǎn)的坐標(biāo)為(x2,y2).則x2+2=
13 |
2 |
∴x2=
9 |
2 |
9 |
2 |
9 |
2 |
9 |
2 |
9 |
2 |
9 |
2 |
17 |
2 |
1 |
2 |
17 |
2 |
1 |
2 |
1 |
2 |
13 |
2 |
13 |
2 |
9 |
2 |
9 |
2 |