If \( x^{4} y^{2}-14 x^{2} y+49=0 \), prove that \( : y \propto \frac{1}{x^{2}} \)

To show that \( y \propto \frac{1}{x^{2}} \) from the equation \( x^{4} y^{2}-14 x^{2} y+49=0 \), we start by treating this as a quadratic equation in terms of \( y \): \[ x^{4} y^{2} - 14 x^{2} y + 49 = 0. \] Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = x^4 \), \( b = -14x^2 \), and \( c = 49 \), we have: \[ y = \frac{14x^{2} \pm \sqrt{(-14x^{2})^2 - 4(x^{4})(49)}}{2(x^{4})}. \] Calculating the discriminant: \[ (-14x^{2})^2 = 196x^{4}, \quad 4(x^{4})(49) = 196x^{4} \quad \Rightarrow \quad b^2 - 4ac = 196x^{4} - 196x^{4} = 0. \] This shows that the quadratic has a double root, thus: \[ y = \frac{14x^{2}}{2x^{4}} = \frac{7}{x^{2}}. \] This equation indicates that \( y \) is inversely proportional to \( x^{2} \), so we conclude that \( y \propto \frac{1}{x^{2}} \).