Theo đề, ta có: \(x_1y_1=x_2y_2\)
\(\Leftrightarrow2y_1=3y_2\)
hay \(\dfrac{y_1}{3}=\dfrac{y_2}{2}\)
Đặt \(\dfrac{y_1}{3}=\dfrac{y_2}{2}=k\)
\(\Leftrightarrow y_1=3k;y_2=2k\)
Ta có: \(y_1^2+y_2^2=52\)
\(\Leftrightarrow13k^2=52\)
\(\Leftrightarrow k^2=4\)
TH1: k=2
=>y1=6; y2=4
TH2: k=-2
=>y1=-6; y2=-4