An elliptical field has the equation of its boundary $x^2+3y^2=3$ with A at an end of its major axis.A tower stands vertically at A and from the points B,C on the boundary the angles of the elevation of the top of the tower are found to be $\frac{\pi}{4}$ each.If $A,B,C$ are on the same side of the minor axis and $BC$ subtends a right angle at the center of the ellipse then find the height of the tower.
How should i approach this problem,I thought over it but no constructive ideas came.
This is not a formal argument, just what I could construct in ten minutes.
WLOG, assume we are working on the right hand side of the ellipse, since working on one side translates to working on the other side.
Claim: $AC = AB$
If the angles of elevation are to be equal, then the base lengths must be the same, since they share the same height. If the bases were not equal, then one of the triangles would have a greater angle of elevation than the other since one would have a point further away, resulting in differing angles of depression. This yields a contradiction, so the bases must be equal.
Since the bases from $A$ are equal in length, then they must have equal $x$ co-ordinates. The $y$ co-ordinates are negatives of each other, since they lie on opposite sides of the ellipse. This results in the x-axis bisecting $\angle BOC$. Since $\angle BOC$ is defined to be a right angle, the angle between the x-axis and the points $B$, $C$, are $\frac{\pi}{2}$. This means that the points lie on the lines y=x and y=-x.
The points of intersection between the ellipse, y=x and y=-x are $(\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2})$ and $(\frac{\sqrt{3}}{2},\frac{-\sqrt{3}}{2})$. (recall we are working on the RHS of the ellipse)
The length OC is easily calculated to be $\sqrt{\frac{3}{2}}$ units.
Using the distance formula, the length of $AC$ turns out to be the same as the length of $OC$. Since the angle of elevation is $\frac{\pi}{2}$, the triangle $TCA$ (where $T$ is the top of the tower) is a right angled (by definition) isosceles triangle (by computing the remaining angle). This means the base $CA$ and the height of the tower $AT$ are the same. Therefore, the height of the tower is $\sqrt{\frac{3}{2}}$ units