[题目]求下列曲线在所示点处的切线与法平面:-|||-=a(sin )^2t =bsin tcos t,z=c(cos )^2t, 在点 =dfrac (pi )(4).

步骤 1：计算曲线在 $t=\dfrac {\pi }{4}$ 处的坐标
根据给定的参数方程，当 $t=\dfrac {\pi }{4}$ 时，我们有：
$x=a{\sin }^{2}(\dfrac {\pi }{4})=a(\dfrac {\sqrt{2}}{2})^{2}=\dfrac {a}{2}$
$y=b\sin (\dfrac {\pi }{4})\cos (\dfrac {\pi }{4})=b(\dfrac {\sqrt{2}}{2})(\dfrac {\sqrt{2}}{2})=\dfrac {b}{2}$
$z=c{\cos }^{2}(\dfrac {\pi }{4})=c(\dfrac {\sqrt{2}}{2})^{2}=\dfrac {c}{2}$
步骤 2：计算曲线在 $t=\dfrac {\pi }{4}$ 处的切线方向向量
切线方向向量由参数方程的导数给出，即：
$x'(t)=2a\sin t\cos t$
$y'(t)=b(\cos^{2}t-\sin^{2}t)$
$z'(t)=-2c\sin t\cos t$
当 $t=\dfrac {\pi }{4}$ 时，我们有：
$x'(\dfrac {\pi }{4})=2a\sin (\dfrac {\pi }{4})\cos (\dfrac {\pi }{4})=2a(\dfrac {\sqrt{2}}{2})(\dfrac {\sqrt{2}}{2})=a$
$y'(\dfrac {\pi }{4})=b(\cos^{2}(\dfrac {\pi }{4})-\sin^{2}(\dfrac {\pi }{4}))=b(\dfrac {1}{2}-\dfrac {1}{2})=0$
$z'(\dfrac {\pi }{4})=-2c\sin (\dfrac {\pi }{4})\cos (\dfrac {\pi }{4})=-2c(\dfrac {\sqrt{2}}{2})(\dfrac {\sqrt{2}}{2})=-c$
因此，切线方向向量为 $(a,0,-c)$。
步骤 3：写出切线方程
切线方程为：
$\dfrac {x-\dfrac {a}{2}}{a}=\dfrac {y-\dfrac {b}{2}}{0}=\dfrac {z-\dfrac {c}{2}}{-c}$
步骤 4：计算法平面方程
法平面方程由切线方向向量给出，即：
$a(x-\dfrac {a}{2})+0(y-\dfrac {b}{2})-c(z-\dfrac {c}{2})=0$
整理得：
$ax-cz=\dfrac {1}{2}({a}^{2}-{c}^{2})$