Q.Find the angle between the curves $2y^2 - 9x = 0$, $3x^2 + 4y = 0$ (in the $4^{th}$ quadrant).

Find the intersection point of the two curves in the 4th quadrant, compute each curve's slope there, then apply the angle-between-curves (tangent lines) formula.

Curves: $2y^2-9x=0$ ... (1) and $3x^2+4y=0$ ... (2)

Find the intersection point in the 4th quadrant ($x>0,y<0$):

From (2): $y=-\dfrac{3x^2}{4}$. Substitute into (1):

$2\left(-\dfrac{3x^2}{4}\right)^2 - 9x = 0 \implies 2\cdot\dfrac{9x^4}{16} - 9x=0 \implies \dfrac{9x^4}{8}=9x \implies x^3=8 \implies x=2$

(rejecting $x=0$, the origin, which is not in the 4th quadrant)

$y = -\dfrac{3(2)^2}{4} = -3$

So the point is $(2,-3)$, which lies in the 4th quadrant.

Slope of curve (1) at $(2,-3)$: differentiate $2y^2-9x=0$ implicitly: $4y\dfrac{dy}{dx}-9=0 \implies \dfrac{dy}{dx}=\dfrac{9}{4y}$

At $y=-3$: $m_1 = \dfrac{9}{4(-3)} = -\dfrac{3}{4}$

…