Give an upper bound for $x^2 + y^2$ for $(x, y)$ on the ellipse $(x - y - a)^2 + y^2 = 1$ without using Lagrange multipliers Let $a \in \mathbb{R}$ and consider the ellipse $(x - y - a)^2 + y^2 = 1$. My goal is to maximize (or at least give an upper bound for) $x^2 + y^2$ on this ellipse. This can be solved with Lagrange multipliers however the calculations are quite messy and hard to simplify (mostly due to $a$ being any real number). 
Are there any easier alternative ways that avoid using Lagrange multipliers, perhaps at the cost of showing instead there is a $C$ depending on $a$ such that $x^2 + y^2 \leq C$ for all $(x, y)$ on the ellipse?
 A: This doesn't avoid Lagrange multipliers, but you can let $z=x-y$ and then 
try to find an upper bound for $(z+y)^2+y^2$ on the circle $(z-a)^2+y^2=1$:
1) If we maximize $(z+y)^2$ on the circle, we get 
$\hspace{.3 in}2(z+y)=\lambda\cdot2(z-a)$ and $2(z+y)=\lambda\cdot2y$, so the maximum occurs for $z=a+y$,
and then $(z-a)^2+y^2=1\implies y=\pm\frac{1}{\sqrt{2}}$ and $z=a\pm\frac{1}{\sqrt{2}}$.
Therefore $(z+y)^2+y^2\le (a+\sqrt{2})^2+y^2\le a^2+2\sqrt{2}a+2+1=\color{green}{a^2+2\sqrt{2}a+3}$
A: Suggestions: If you let $F(x,y)=(x-y-a)^2+y^2$ then $x^2+y^2$ attains the max on $F=1$ at a point which is the furthest away from the origin, i.e. where the gradient is parallel to $(x,y)$  so you may solve $\nabla F \cdot (-y,x)=0$ simultaneously with $F(x,y)-1=0$ to obtain the point, whence the value.
(This procedure hides the Lagrange multiplier under the rug).
Later edit (also gives a sort of general method): We may write 
 $$   F(x,y)=\left( \begin{matrix} x-a & y \end{matrix} \right) \left( \begin{matrix} 1  & -1 \\ -1 & 2 \end{matrix} \right) \left( \begin{matrix} x-a \\ y \end{matrix} \right) \geq \lambda_{\rm min}\left( (x-a)^2 + y^2 \right)$$
where $\lambda_{\rm min}=\frac{2}{3+\sqrt{5}}=\left(\frac{2}{1+\sqrt{5}}\right)^2$ is the minimal eval of the matrix. The center of the ellipse $F=1$ is $(a,0)$ and the maximal distance from the center is bounded by $r=1/\sqrt{\lambda_{\rm min}}$. So the maximal distance from the origin is  $|(a,0)|+r$ and therefore,
$$ \max\{x^2+y^2 : F\leq 1\} \leq \left(|a|+ \frac{1+\sqrt{5}}{2}\right)^2= a^2 + (1+\sqrt{5})|a| + \frac{3+\sqrt{5}}{2}$$
