8. On the Oy axis, find the point through which two perpendicular to each other tangents to the graph of the function $y=1-2 x-x^{2}$ pass, and write the equations of these tangents.

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# Solution:

$y=1-2 x-x^{2}, M\left(0 ; y_{0}\right)$. Equation of the tangent: $y=y_{0}+k x$.

The equation $1-2 x-x^{2}=y_{0}+k x$, or $x^{2}+(k+2) x+y_{0}-1=0$, has a unique solution if $D=k^{2}+4 k+8-4 y_{0}=0$. The two values of $k$ found from this equation must satisfy the condition of perpendicularity $k_{1} \cdot k_{2}=-1$. Since $k_{1} \cdot k_{2}=8-4 y_{0}=-1, y_{0}=9 / 4$. From the equation $k^{2}+4 k-1=0$ we find $k_{1,2}=-2 \pm \sqrt{5}$.

Answer: $M(0 ; 9 / 4)$, equations of the tangents: $y=9 / 4+(-2 \pm \sqrt{5}) x$.