On an ellipse!

Given the ellipse ${\rm C}: x^2-xy+y^2=\frac{3}{4}$

1. Find its area.
2. Find the points of the ellipse that a horizontal tangent attaches them as well as the points that a vertical tangent attaches them.
3. Which are the most far most and most close points of the ellipse?

Solution

1. First of all we note that ellipse is rotated to $\frac{\pi}{4}$ degrees since $\tan 2\varphi = \frac{2b}{c-a}=\infty$. Applying the rotation (either using eigenvalues or by using the classical way) our ellipse comes to form:
   
   $${\rm C}:\frac{X^2}{\left ( \sqrt{\frac{3}{2}} \right )^2} + \frac{Y^2}{\left ( \sqrt{\frac{1}{2}} \right )^2}=1$$
   
   Hence $a= \sqrt{\frac{3}{2}}, \; \beta = \sqrt{\frac{1}{2}}$. The area of an ellipse is given by the formual ${\rm E}=\pi a \beta$. Thus:
   
   $${\rm E}= \pi a \beta = \sqrt{\frac{3}{2}}\sqrt{\frac{1}{2}}\pi = \frac{\sqrt{3} \pi}{2}$$


2. Let us consider the function $f(x, y)= x^2-xy+y^2-\frac{3}{4}$. Then the nabla of this function is:
   
   $$\nabla f = \left ( f_x (x, y), \; f_y (x, y) \right )= \left ( 2x-y , -x+2y \right )$$
   
   Since we want the tangent to be perpendicular to the $x'x$ axis , we do nothing else by just equating $f_y$ to zero. Thus if $(x_0, y_0)$ is such a point then:
   
   $$\left\{\begin{matrix}
   x_0=2y_0 & \\
    x_0^2 -x_0 y_0 +y_0^2 = \frac{3}{4}&
   \end{matrix}\right. \Rightarrow \left\{\begin{matrix}
   x_0= 1 \; \text{or} \; -1 & \\
   y_0= \frac{1}{2} \; \text{or} \; -\frac{1}{2} &
   \end{matrix}\right.$$
   
   Thus the points that the tangent to each of them is perpendicular to the $x'x$ axis are ${\rm A} \left ( -1, - \frac{1}{2} \right ) $ and ${\rm B} \left ( 1,  \frac{1}{2} \right ) $.
   
   Similarly, since we want the tangent to be perpendicular to the $y'y$ axis , we do nothing else by just equating $f_x$ to zero. Doing so, we find the other points which are ${\rm A'}\left ( -\frac{1}{2}, -1 \right )$ and ${\rm B'}\left ( \frac{1}{2}, 1 \right )$.


3. The foci of the ellipse are the points $\Gamma' \left ( - \frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}} \right ), \; \Gamma \left ( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right )$. The center of the ellipse is , obviously, the origin. Thus to find the far most points of the ellipse we solve the system that is consisted of the equation of the ellipse and the line $y=x$. We are ommiting the calculations because they are easy. This results in $2$ points.
   
   $${\rm A'} \left ( - \frac{\sqrt{3}}{2}, \; - \frac{\sqrt{3}}{2} \right ), \quad {\rm A} \left ( \frac{\sqrt{3}}{2}, \; \frac{\sqrt{3}}{2} \right )$$
   
   On the other hand , to determine the close most points of the ellipse we solve the system that is consisted of the equation of the ellipse and the line $y=-x$. Again, since calculations are easy we ommit them. This results in the following two points.
   
   $${\rm B'} \left ( - \frac{1}{2}, \;  \frac{1}{2} \right ), \quad {\rm B} \left ( \frac{1}{2}, \; -\frac{1}{2} \right )$$