Question 36: Find the values of the parameter m so that the graph of the function: \(y=x^{4}-2 m^{2} x^{2}+1\) has three extreme points which are three vertices of an isosceles right triangle?
\(\begin{array}{l} y^{\prime}=4 x^{3}-4 m^{2} x \\ y^{\prime}=0 \Leftrightarrow 4 x\left(x^ {2}-m^{2}\right)=0 \end{array}\)
\(\text { Function with } 3 \text { extreme point } \Leftrightarrow m \neq 0\)
\(\text { Then } 3 \text { the extreme point of the function graph is : } A(0 ; 1), B\left(m ; 1-m^{4}\right), C\left (-m ; 1-m^{4}\right)\)
\(\text { Due to symmetry, we have } \Delta ABC \text { isosceles at the vertex } A \text { . }\)
\(\text { So } \Delta ABC \text { can only be isosceles at the vertex } A \Leftrightarrow \overrightarrow{AB} \cdot \overrightarrow{AC}=0 \Leftrightarrow-m^{2}+m^{ 8}=0 \Leftrightarrow\left[\begin{array}{l}m=0\\m=\pm1\end{array}\right\text{}\)
\(\text { Combining the conditions we have: } m=\pm 1 \text { (satisfies). }\)
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