8. On the line $3 x+4 y=2$ find the point through which two perpendicular to each other tangents to the graph of the function $y=x^{2}$ pass, and write the equations of these tangents.

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# Solution:

$y=x^{2}, 3 x+4 y=2, M\left(x_{0} ; y_{0}\right)$. The equation of the tangent line: $y=y_{0}+k\left(x-x_{0}\right)$.

The equation $x^{2}=y_{0}+k\left(x-x_{0}\right)$, or $x^{2}-k x+k x_{0}-y_{0}=0$, has a unique solution if $D=k^{2}-4 k x_{0}+4 y_{0}=0$. The two values of $k$ found from this equation must satisfy the condition of perpendicularity $k_{1} \cdot k_{2}=-1$. Since $k_{1} \cdot k_{2}=4 y_{0}, y_{0}=-1 / 4$. From the equation of the line $x_{0}=\left(2-4 y_{0}\right) / 3=(2+1) / 3=1$. From $k^{2}-4 k-1=0$ we find $k_{1,2}=2 \pm \sqrt{5}$.

Answer: $M(1 ;-1 / 4)$, equations of the tangents: $y=-1 / 4+(2 \pm \sqrt{5})(x-1)$.