9. (1 mark) Let $x_{1}, y_{1}, x_{2}, y_{2}$ be real numbers satisfying the equations $x_{1}^{2}+5 x_{2}^{2}=10$, $x_{2} y_{1}-x_{1} y_{2}=5$ and $x_{1} y_{1}+5 x_{2} y_{2}=\sqrt{105}$. Find the value of $y_{1}^{2}+5 y_{2}^{2}$.
(1 mark) Let $x_{1}, y_{1}, x_{2}, y_{2}$ be real numbers satisfying the equations $x_{1}^{2}+5 x_{2}^{2}=10$, $x_{2} y_{1}-x_{1} y_{2}=5$ and $x_{1} y_{1}+5 x_{2} y_{2}=\sqrt{105}$. Find the value of $y_{1}^{2}+5 y_{2}^{2}$.
9. $\left(x_{1}^{2}+5 x_{2}^{2}\right)\left(y_{1}^{2}+5 y_{2}^{2}\right)=\left(x_{1} y_{1}+5 x_{2} y_{2}\right)^{2}+5\left(x_{2} y_{1}-x_{1} y_{2}\right)^{2}$, thus $10\left(y_{1}^{2}+5 y_{2}^{2}\right)=105+5 \times 5^{2}=230$, or
$$
y_{1}^{2}+5 y_{2}^{2}=\underline{\underline{23}} .
$$

Alternate Solution

From $x_{1}^{2}+5 x_{2}^{2}=10$, we may have let $x_{1}=\sqrt{10} \cos \alpha, x_{2}=\sqrt{2} \sin \alpha$. Similarly if we let $y_{1}=r \cos \beta$ and $y_{2}=\frac{r}{\sqrt{5}} \sin \beta$, then $y_{1}^{2}+5 y_{2}^{2}=r^{2}$. From $x_{2} y_{1}-x_{1} y_{2}=5$ and $x_{1} y_{1}+5 x_{2} y_{2}=\sqrt{105}$, we have
$$
\left\{\begin{array}{l}
\sqrt{2} \sin \alpha \cdot r \cos \beta-\sqrt{10} \cos \alpha \cdot \frac{r}{\sqrt{5}} \sin \beta=\sqrt{2} r \sin (\alpha+\beta)=5 \\
\sqrt{10} \cos \alpha \cdot r \cos \beta+5 \sqrt{2} \sin \alpha \cdot \frac{r}{\sqrt{5}} \sin \beta=\sqrt{10} r \cos (\alpha-\beta)=\sqrt{105}
\end{array}\right. \text {. }
$$

This implies $r^{2}=\left(\frac{5}{\sqrt{2}}\right)^{2}+\left(\frac{\sqrt{105}}{\sqrt{10}}\right)^{2}=\underline{\underline{23}}$
Alternate Solution 2

Apparently any special case should work. Try $x_{1}=\sqrt{10}, x_{2}=0$. From $x_{2} y_{1}-x_{1} y_{2}=5$, get $y_{2}=-\frac{5}{\sqrt{10}}$. From $x_{1} y_{1}+5 x_{2} y_{2}=\sqrt{105}$, get $y_{1}=\frac{\sqrt{105}}{\sqrt{10}}$. Hence $y_{1}^{2}+5 y_{2}^{2}=\frac{105}{10}+\frac{5 \cdot 25}{10}=\underline{\underline{23}}$.