No Integration Necessary (Part 2)

Okay, I went against the recommendation given in the question because I simply love calculus. For the given ellipse, the semi-major axis is $a=5$ and the semi-minor axis is $b=2$ Let the equation of the line be $y=mx$ since it passes through the origin. Solving this and the equation for the ellipse for x gives you the x coordinate of the intersection point of the line and the ellipse. $x=\frac{10}{\sqrt{25m^2+4}}=k$ Now, the line cuts the quarter ellipse into two halves of equal area. Since the area under the quarter ellipse is $\frac{\pi ab}{4}$ , each of the area cut by the line is $\frac{\pi ab}{8} or \frac{5\pi}{4}$ Now solve the equation $\int_0^{k} mx dx + \int_k^{5} \frac{2}{5} \sqrt{25-x^2} dx = \frac{5\pi}{4}$ Do this, substitute for k and get the equation: $\frac{\pi}{2} - \sin^{-1}{\frac{2}{\sqrt{25m^2 + 4}}} = \frac{\pi}{4}$ or, $\cos^{-1}{\frac{2}{\sqrt{25m^2 + 4}}} = \frac{\pi}{4}$ or, $\tan^{-1}{\frac{5m}{2}} = \frac{\pi}{4}$ or, $\frac{5m}{2} = 1$ or, $m = \frac{2}{5}$ Finally, the angle made by the line with the x-axis is $\tan^{-1}{m} = \tan^{-1}{\frac{2}{5}} = 0.38 rad = \boxed{21.8^{\circ}}.$

Consider the circle circumscribing the ellipse. It will have a radius = semi major axis = 5 This circle is 'tilted' about the x axis to get the given ellipse. The tilt is such that the vertical radius 5 is now seen as 2 units. So a line with 45° inclination dividing the quarter circle in two equal areas will now have an inclination $\arctan \frac{2}{5} = 21.80$