Answer: \(k=7-5\sqrt2\).

Answer: \(k=7-5\sqrt2\).

Since \(y=kx^2\),

\(\displaystyle k=\frac{y}{x^2}\).

Here \(x=1+\sqrt2\) and \(y=1-\sqrt2\), so

\(\displaystyle k=\frac{1-\sqrt2}{(1+\sqrt2)^2}\).

Now

\((1+\sqrt2)^2=1+2\sqrt2+2=3+2\sqrt2\).

Therefore

\(\displaystyle k=\frac{1-\sqrt2}{3+2\sqrt2}\).

Rationalise the denominator using \(3-2\sqrt2\):

\(\displaystyle k=\frac{(1-\sqrt2)(3-2\sqrt2)}{(3+2\sqrt2)(3-2\sqrt2)}\).

The denominator is \(9-8=1\). The numerator is

\(3-2\sqrt2-3\sqrt2+4=7-5\sqrt2\).

Hence \(k=7-5\sqrt2\).