Let's rewrite the equation in form
2x=y(1−x)
Hence
y=1−x2x
The equation of the tangent
yt=y′(x0)(x−x0)+y(x0)
y′=(1−x)22(1−x)+2x=(1−x)22
Since the tangent is parallel to the line
2x+y=0
y′(x0)=−2⟺(1−x)22=−2
The last equation has no real roots hence there are no tangents to this curve which are parallel to this line.
The equation of the normal
yn=−y′(x0)1(x−x0)+y(x0)
Since the normal is parallel to the line
2x+y=0
−y′(x0)1=−2⟺−2(1−x2)=−2
(1−x)2=4
x1=3;x2=−1
Two normals to the curve have the equation
yn1=−2(x−3)−3=−2x+3
yn2=−2(x+1)−1=−2x−3
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