91Ó°ÊÓ

To find the point(s) of intersection, we need to solve the system of equations composed of the circle's equation and the parabola's equation. Solve for the intersection points of \(x^2 + y^2 = 16\) and \(y^2 = 6x\). From equation \(y^2 = 6x\), we can write \(x = \frac{1}{6}y^2\). Substitute this in the circle's equation to get: \(\frac{1}{6}y^4 + y^2 = 16\) Solve for y: \[\frac{1}{6}y^4 + y^2 - 16 = 0\] Using a numerical approximation method or otherwise finding the roots of this equation, we find that \(y \approx \pm 2\) . Now, we have two potential intersection points. Let's find the corresponding x values for each. If \(y = 2\), \(x = \frac{1}{6}(4) = \frac{2}{3}\). If \(y = -2\), \(x = \frac{1}{6}(4) = \frac{2}{3}\). Thus, the points of intersection are \((\frac{2}{3}, 2)\) and \((\frac{2}{3}, -2)\).

To compute the gradient of the circle at the point of interscetion, we can differentiate the equation \(x^2 + y^2 = 16\) implicitly with respect to x. \[\frac{d}{dx} (x^2 + y^2) = \frac{d}{dx}(16)\] \[2x + 2y \frac{dy}{dx} = 0\] Now, solve for \(\frac{dy}{dx}\): \[\frac{dy}{dx}_{circle} = -\frac{x}{y}\] Similarly, differentiate the equation of the parabola \(y^2 = 6x\) implicitly with respect to x: \[\frac{d}{dx}(y^2) = \frac{d}{dx}(6x)\] \[2y \frac{dy}{dx} = 6\] Now, solve for \(\frac{dy}{dx}\): \[\frac{dy}{dx}_{parabola} = \frac{3}{y}\]

We now have the gradients of both the circle and the parabola at the intersection point. In this case, we have two intersection points, so we need to compute the angle at each one. Recall that the formula for the angle \(\theta\) between two curves with gradients \(m_1\) and \(m_2\) is \[\tan(\theta) = \frac{m_1 - m_2}{1 + m_1m_2}\] For intersection point \((\frac{2}{3}, 2)\), the gradients of the circle and parabola are: \(\frac{dy}{dx}_{circle} = -\frac{\frac{2}{3}}{2} = -\frac{1}{3}\) \(\frac{dy}{dx}_{parabola} = \frac{3}{2}\) Now, compute the angle \(\theta\): \[\tan(\theta) = \frac{-\frac{1}{3} - \frac{3}{2}}{1 -\frac{1}{3} \cdot \frac{3}{2}} = \frac{5}{7}\] So, the angle at the point \((\frac{2}{3}, 2)\) is given by \(\theta = \arctan(\frac{5}{7})\). Similarly, for intersection point \((\frac{2}{3}, -2)\), the gradients of the circle and parabola are: \(\frac{dy}{dx}_{circle} = -\frac{\frac{2}{3}}{-2} = \frac{1}{3}\) \(\frac{dy}{dx}_{parabola} = -\frac{3}{2}\) Now, compute the angle \(\theta\): \[\tan(\theta) = \frac{\frac{1}{3} + \frac{3}{2}}{1 +\frac{1}{3} \cdot -\frac{3}{2}} = \frac{5}{7}\] So, the angle at the point \((\frac{2}{3}, -2)\) is given by \(\theta = \arctan(\frac{5}{7})\).

The angle between the curves \(x^2 + y^2 = 16\) and \(y^2 = 6x\) at both intersection points \((\frac{2}{3}, 2)\) and \((\frac{2}{3}, -2)\) is given by \(\theta = \arctan(\frac{5}{7})\).