Curves That Cut at Right Angles (Application of Derivatives)

This lesson proves a result from application of derivatives: the curves $2x = y^2$ and $2xy = k$ cut at right angles exactly when $k^2 = 8$. We find their point of intersection, compute the slope of each tangent there, and use the condition that perpendicular tangents have slopes whose product is $-1$.

The two curves

We are given two curves and label them:

$2x = y^2 \quad (1)$

$2xy = k \quad (2)$

From $(1)$ we get $x = \tfrac{y^2}{2}$, and from $(2)$ we get $x = \tfrac{k}{2y}$.

Finding the point of intersection

At a common point both expressions for $x$ are equal, so

$\frac{y^2}{2} = \frac{k}{2y}.$

Cross-multiplying gives $2y^3 = 2k$, hence

$y^3 = k \quad\Longrightarrow\quad y = k^{1/3}.$

Substituting back into $(1)$:

$x = \frac{y^2}{2} = \frac{\left(k^{1/3}\right)^2}{2} = \frac{k^{2/3}}{2}.$

So the curves meet at

$\left( \frac{k^{2/3}}{2},\ k^{1/3} \right).$

Slope of the tangent to the first curve

Differentiate $(1)$, $2x = y^2$, with respect to $x$:

$2 = 2y\,\frac{dy}{dx} \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{1}{y}.$

At the intersection point $y = k^{1/3}$, so the slope of the first tangent is

$m_1 = \frac{1}{k^{1/3}} = k^{-1/3}.$

Slope of the tangent to the second curve

Differentiate $(2)$, $2xy = k$, using the product rule:

$2\left( x\,\frac{dy}{dx} + y \right) = 0 \quad\Longrightarrow\quad x\,\frac{dy}{dx} + y = 0,$

so

$\frac{dy}{dx} = -\frac{y}{x}.$

Substituting $y = k^{1/3}$ and $x = \tfrac{k^{2/3}}{2}$:

$m_2 = -\frac{k^{1/3}}{\tfrac{k^{2/3}}{2}} = -2\,k^{1/3 - 2/3} = -2\,k^{-1/3}.$

Applying the right-angle condition

The curves cut at right angles, so their tangents are perpendicular and

$m_1 \cdot m_2 = -1.$

Substituting the slopes:

$k^{-1/3}\cdot\left(-2\,k^{-1/3}\right) = -1 \quad\Longrightarrow\quad -2\,k^{-2/3} = -1.$

Therefore

$k^{-2/3} = \frac{1}{2}.$

Simplifying to the required result

Taking the cube of both sides removes the fractional power:

$\left(k^{-2/3}\right)^3 = \left(\frac{1}{2}\right)^3 \quad\Longrightarrow\quad k^{-2} = \frac{1}{8}.$

Hence

$k^2 = 8.$

So the curves cut at right angles precisely when $k^2 = 8$, as required.

Key takeaways

• Two curves cut at right angles when their tangents at the intersection point are perpendicular, i.e. $m_1 m_2 = -1$.
• Use implicit differentiation to find $\tfrac{dy}{dx}$ for each curve, then evaluate it at the common point.
• For these curves the slopes are $m_1 = k^{-1/3}$ and $m_2 = -2k^{-1/3}$, and the perpendicularity condition simplifies to $k^2 = 8$.