I am having an issue where I am getting two different answers and I am not able to figure out where is the mistake. Both methods seems correct to me.
Find the eccentric angles of the extremities of the latus recta of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The centre of the ellipse is at origin.
Any point in an ellipse P is $(acos(θ), bsin(θ))$, the coordinates of one end of the latus recta are $(ae, \frac{b^2}{a})$. therefore, $(acos(θ), bsin(θ)) = (ae, \frac{b^2}{a})$ thus, $cos(θ)=e$ and $sin(θ)=\frac{b}{a}$, $\implies tan(θ)=\frac{b}{ae}$,hence $θ=arctan(\frac{b}{ae})$
For a traingle with perpendicular, $P=\frac{b^2}{a}$, and base, $B=ae$, $tan(θ) = \frac{P}{B} = \frac{\frac{b^2}{a}}{ae} = \frac{b^2}{a^2e}$. Hence, $θ=arctan(\frac{b^2}{a^2e})$
does this mean $\frac{b}{a} = 1$? which is only true for circles.
What am I doing wrong please help...