I have a question that says that:
A line $x+2y+4 = 0$ cutting the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ in points whose eccentric angles are $30^\circ$ and $60^\circ$ subtends a right angle at the origin, then the equation of the ellipse is:
The answer is given as $$\frac{x^2}{16}+\frac{y^2}{4} = 1$$
Now, the solution the book provides was by first considering the points as $(a \cos\theta,b \sin\theta)$, and using the values of 2 given angles, and then equating it with the slope of the line, and thus they found the relation that $a = 2b$ and then what they did was homogenise the equation of the line with the ellipse and set the coeffcient of $x^2 + y^2$ = 0 which is the condition for subtending $90^\circ$ at the center, and they get $b^2 = 4 $
This solution is perfectly fine, and I'm able to understand it, but what I tried was, since the line cuts the ellipse at the given points, I took the points as $(a \cos\theta,b \sin\theta)$ and then I put the values of given angles. Then I put the points into the equation of the line as this should satisfy the equation of the line as these points lie on it.
But this yields a different value of $b$, and as far as I can tell neither of the calculations have had a calculation mistake. I dont know why my solution is wrong or what I'm missing here.
Best Answer
As stated, the problem has no solution. The intersection points are: $$ A=(a\cos30°,a\sin30°)=\bigg({\sqrt3\over2}a,{1\over2}b\bigg); \quad B=(a\cos60°,b\sin60°)=\bigg({1\over2}a,{\sqrt3\over2}b\bigg); $$ but $\angle AOB$ cannot be a right angle, because it is easy to compute: $$ \cos\angle AOB={A\cdot B\over|A|\ |B|}= {\sqrt3(a^2+b^2)\over2\sqrt{3a^2+b^2}\sqrt{a^2+3b^2}}\ne0. $$
EDIT.
The given solution uses the eccentric angles only to find that $a=2b$. But this is not specific of the angles $30°$ and $60°$: in fact the points where the line intersects the ellipse given as solution correspond to eccentric angles of $180°$ and $270°$.
Your method is perfectly fine, but it cannot find a valid solution in this case because the given line lies in the second, third and fourth quadrant, whereas the points corresponding to eccentric angles of $30°$ and $60°$ lie in the first quadrant.