Question-50
Find the points on the ellipse \(\displaystyle 4x^{2} +y^{2} =4\) that are farthest away from the point \(\displaystyle ( 1,0)\).
Let \(\displaystyle ( x,y)\) be a point on the ellipse. The square of the distance of this point from \(\displaystyle ( 1,0)\) is:
\[ \begin{equation*} d^{2} =( 1-x)^{2} +y^{2} \end{equation*} \]
Since \(\displaystyle ( x,y)\) lies on the ellipse, this turns into:
\[ \begin{equation*} \begin{aligned} d^{2} & =( 1-x)^{2} +4-4x^{2}\\ & =1+x^{2} -2x+4-4x^{2}\\ & =5-2x-3x^{2} \end{aligned} \end{equation*} \]
The optimization problem is therefore:
\[ \begin{equation*} \underset{x}{\max} \ \ 5-2x-3x^{2} \end{equation*} \]
subject to \(\displaystyle x\in [ -1,1]\). This is a constrained optimization problem. But let us not worry about that right now and try to treat this as though it were unconstrained. The parabola is concave downward and has a global maximum at \(\displaystyle x=-\frac{1}{3}\). Thankfully, this value is inside the range \(\displaystyle [ -1,1]\). We can get \(\displaystyle y=\pm \sqrt{4-\frac{4}{9}} =\pm \frac{4}{3}\sqrt{2}\). This gives us two points:
\[ \begin{equation*} \boxed{\left(\frac{-1}{3} ,\frac{4\sqrt{2}}{3}\right) ,\ \left(\frac{-1}{3} ,\frac{-4\sqrt{2}}{3}\right)} \end{equation*} \]