The curve x = y^2 and xy = k cut orthogonally, then k^2 is equal to:

The correct answer is Option (4) → \(\displaystyle\frac{1}{8}\)

Two curves cut orthogonally when their tangents at the point of intersection are perpendicular, that is \(m_1 m_2 = -1\). Let (x, y) be the point of intersection.

For \(x = y^2\), differentiating with respect to x:

    \( \displaystyle 1 = 2y\frac{dy}{dx},\quad m_1 = \frac{1}{2y} \)

For \(xy = k\):

    \( \displaystyle y + x\frac{dy}{dx} = 0,\quad m_2 = -\frac{y}{x} \)

Then

    \( \displaystyle m_1 m_2 = \frac{1}{2y} \times \left(-\frac{y}{x}\right) = -\frac{1}{2x} \)

    \( \displaystyle -\frac{1}{2x} = -1,\quad x = \frac{1}{2} \)

The point lies on \(x = y^2\), so \(y^2 = \frac{1}{2}\). Hence

    \( \displaystyle k^2 = x^2y^2 = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \)

Tip: The condition \(8k^2 = 1\) for the curves \(x = y^2\) and \(xy = k\) to cut at right angles is a standard result worth remembering.