# 23) Prove that the curves $x = y^2$ and xy = k cut at right angles* $if \: \: 8k ^ 2 = 1.$

Let suppose, Curve $x = y^2$ and xy = k cut at the right  angle
then the slope of their tangent also cut  at the right angle
means,
$\left ( \frac{dy}{dx} \right )_a \times \left ( \frac{dy}{dx} \right )_b = -1$                                                                   -(i)
$2y\left ( \frac{dy}{dx} \right )_a = 1 \Rightarrow \left ( \frac{dy}{dx} \right )_a = \frac{1}{2y}$
$\left ( \frac{dy}{dx} \right )_b = \frac{-k}{x^2}$
Now these values in equation (i)
$\frac{1}{2y} \times \frac{-k}{x^2} = -1\\ -k = -2yx^2\\ k =2(xy)(x)\\ k = 2k(k^{\frac{2}{3}}) \ \ \ \ \left ( x = y^2 \Rightarrow y^2y = k \Rightarrow y = k^{\frac{1}{3}} \ and \ x = k^{\frac{2}{3}} \right ) \\ 2(k^{\frac{2}{3}}) = 1\\ \left ( 2(k^{\frac{2}{3}}) \right )^3 = 1^3\\ 8k^2 = 1$
Hence proved

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