Question

Solve the following equation.
If the equations $ax+by=1,c{x}^2+d{y}^2=1$ have only one solution, prove that $\frac{a^2}{c}+\frac{b^2}{d}=1$ and $x=\frac{a}{x},y=\frac{b}{d}$.

Answer 1

The given equations are
$ax+by=1$ .$\left(1\right)$
$c{x}^2+d{y}^2=1$ $\left(2\right)$
Eliminating y from $\left(1\right)$ and $\left(2\right)$, we obtain
$c{x}^2+d{\left(\frac{1-ax}{b}\right)}^2=1$
or $b^{2}c{x}^2+d-2adx+a^{2}d{x}^2=b^2$
or $(b^{2}c+a^{2}d)x^2-2adx+d-b^2=0$ .$\left(3\right)$
Since the given equations have only one solution, the roots of $\left(3\right)$ must be equal, the condition for which is
$4a^2{d}^2-4(b^{2}c+a^{2}d)(d-b^2)=0$
or $a^2{d}^2-b^{2}cd+b^{4}c-a^2{d}^2=a^2{b}^{2}d=0$
or $b^{4}c+a^2{b}^{2}d-b^{2}cd=0$
or $b^{2}c+a^d-cd=0$
or $\frac{a^2}{c}+\frac{b^2}{d}=1$. ..$\left(4\right)$
Under the condition $\left(4\right)$, the root of $\left(3\right)$ is given by $x+x=$sum
$x=\frac{2ad}{2(b^{2}c+a^{2}d)}=\frac{2ad}{2cd(\frac{b^2}{d}+\frac{a^2}{c})}=\frac{a}{c}$, by $\left(4\right)$
Then $y=\frac{1-ax}{b}=\frac{1-\frac{a^2}{c}}{b}=\frac{\frac{b^2}{d}}{b}$ by $\left(4\right)$ $=\frac{b}{d}$.
Hence the only one solution of the given equations is $x\frac{a}{c},y=\frac{b}{d}$.