Find the angle between the curves $2 y^{2} - 9 x = 0, 3 x^{2} + 4 y = 0$ [in 4th Quadrant]

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Detailed Solution

Given curves

$2 y^{2} - 9 x = 0 \to (1);$

$3 x^{2} + 4 y = 0 \to (2)$

$2 y^{2} = 9 x \Rightarrow x = \frac{2 y^{2}}{9}$

$x$ value sub in equation (2)

$3 (\frac{4 y^{4}}{81}) + 4 y = 0 \Rightarrow 4 y (\frac{y^{3}}{27} + 1) = 0$

$4 y \neq 0 \Rightarrow \frac{y^{3}}{27} + 1 = 0 \Rightarrow y^{3} + 27 = 0$

$y^{3} = - 27 \Rightarrow y^{3} = - 3^{3} \Rightarrow y = - 3$

$∴ x = \frac{2 y^{2}}{9} = \frac{2 (9)}{9} = 2 \Rightarrow P = (2, - 3)$

from (1) $2 y^{2} - 9 x = 0$

diff w.r.t ‘ $x$ ’

$2 (2 y) \frac{d y}{d x} - 9 = 0 \Rightarrow 4 y \frac{d y}{d x} = 9 \Rightarrow \frac{d y}{d x} = \frac{9}{4 y}$

slope $(m) = {(\frac{d y}{d x})}_{P (2, - 3)} = \frac{9}{4 (- 3)}; m_{1} = \frac{- 3}{4}$

from (2) $3 x^{2} + 4 y = 0$

diff w.r.t ‘ $x$ ’

$3 (2 x) + 4 \frac{d y}{d x} = 0 \Rightarrow 4 \frac{d y}{d x} = - 6 x$

$\frac{d y}{d x} = \frac{- 6 x}{4} = \frac{- 3 x}{2}$

slope $(m) = {(\frac{d y}{d x})}_{P (2, - 3)} = \frac{- 3 (2)}{2}$

$m_{2} = - 3$

$\tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{\frac{- 3}{4} + 3}{1 + (\frac{3}{4}) 3}| = \tan θ = |\frac{- 3 + 12}{4 + 9}|$

$θ = \tan^{- 1} (\frac{9}{13})$

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