Area inside an ellipse

Question

Given the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$, $ \ A = (5,0), \ B = (0,4)$; Find point $C$ (with both coordinates positive) on the ellipse, such that the area between AC and the ellipse $(S_1)$ will be equal to the area between BC and the ellipse $(S_2)$. EDIT: I have recently came up with a possible solution (below). Is it valid?

Answer 1

EDIT: I have recently came up with an idea to solve it. It goes like this - we can shrink the ellipse into a circle with radius 4. The same problem in a circle is easy: the coordinate of C can be expressed as (4* cos alpha, 4*sin alpha), and when the areas are equal, alpha is 45 degrees (because the areas will be equal when OC bisects angle BOA, since then both chords and arks BC and AC will be equal). So in a circle with radius 4 C = (4*cos 45, 4*sin 45); To return to the ellipse, we multiply the x axis component of every coordinate by 5/4 (and leave the y coordinates the same). So we get that the two areas will be equal when C = (4*5/4*cos 45, 4*sin 45). Is this solution valid?

Answer 2

HINT:

Let the parametric coordinate of $C$ be $(5\cos\phi,4\sin\phi)$

So, the area$(A_1)$ of $\triangle OAC$ can be calculated

Now, the area $(A_2)$ of the sector $AC$ will be $\displaystyle\int_0^{4\sin\phi}y\ dx$ where $\displaystyle y=+4\cdot\sqrt{1-\frac{x^2}{25}}=+4\cdot\frac{\sqrt{25-x^2}}5$

The area between $AC$ and the ellipse$(S1)=A_2-A_1$