Question

Find the angle of intersection of the curves \[y^2 = 4ax \text { and } x^2 = 4by\] .

 

Answer 1

The given curves are
y² = 4ax         .....(1)
x² = 4by         .....(2)
Solving (1) and (2), we get

\[\left( \frac{x^2}{4b} \right)^2 = 4ax\]

\[ \Rightarrow x^4 - 64a b^2 x = 0\]

\[ \Rightarrow x\left( x^3 - 64a b^2 \right) = 0\]

\[ \Rightarrow x = 0, 4 a^\frac{1}{3} b^\frac{2}{3}\]

When x = 0, y = 0
When \[x = 4 a^\frac{1}{3} b^\frac{2}{3} , y = \frac{\left( 4 a^\frac{1}{3} b^\frac{2}{3} \right)^2}{4b} = 4 a^\frac{2}{3} b^\frac{1}{3}\]

Thus, the given curves intersect at (0, 0) and

\[\left( 4 a^\frac{1}{3} b^\frac{2}{3} , 4 a^\frac{2}{3} b^\frac{1}{3} \right)\]

At (0, 0), the angle between the curves is 90°.
Differentiating (1) with respect to x, we get

\[2y\frac{dy}{dx} = 4a\]

\[ \Rightarrow \frac{dy}{dx} = \frac{2a}{y}\]

\[ \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_\left( 4 a^\frac{1}{3} b^\frac{2}{3} , 4 a^\frac{2}{3} b^\frac{1}{3} \right) = \frac{2a}{4 a^\frac{2}{3} b^\frac{1}{3}} = \frac{1}{2} \left( \frac{a}{b} \right)^\frac{1}{3}\]

Differentiating (2) with respect to x, we get

\[2x = 4b\frac{dy}{dx}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{x}{2b}\]

\[ \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_\left( 4 a^\frac{1}{3} b^\frac{2}{3} , 4 a^\frac{2}{3} b^\frac{1}{3} \right) = \frac{4 a^\frac{1}{3} b^\frac{2}{3}}{2b} = 2 \left( \frac{a}{b} \right)^\frac{1}{3}\]

Let θ be the angle between the two curves. Then,

 

\[\tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|\]

\[ \Rightarrow \tan\theta = \left| \frac{\frac{1}{2} \left( \frac{a}{b} \right)^\frac{1}{3} - 2 \left( \frac{a}{b} \right)^\frac{1}{3}}{1 + \frac{1}{2} \left( \frac{a}{b} \right)^\frac{1}{3} \times 2 \left( \frac{a}{b} \right)^\frac{1}{3}} \right|\]

\[ \Rightarrow \tan\theta = \left| \frac{- \frac{3}{2} \left( \frac{a}{b} \right)^\frac{1}{3}}{1 + \left( \frac{a}{b} \right)^\frac{2}{3}} \right| = \left| \frac{3 a^\frac{1}{3} b^\frac{1}{3}}{2\left( a^\frac{2}{3} + b^\frac{2}{3} \right)} \right|\]

\[ \Rightarrow \theta = \tan^{- 1} \left| \frac{3 \left( ab \right)^\frac{1}{3}}{2\left( a^\frac{2}{3} + b^\frac{2}{3} \right)} \right|\]