Find the maximum and minimum distances from the origin to the curve $5x^3+6xy+5y^2-8=0$
My attempt:
We have to maximise and minimise the following function $x^2+y^2$ with the constraint that $5x^3+6xy+5y^2-8=0$.
Let
$$F(x,y)=x^2+y^2+\lambda(5x^2+6xy+5y^2-8)$$
$$\frac{\delta F(x,y)}{\delta x}=2x+\lambda(10x+6y)$$
and
$$\frac{\delta F(x,y)}{\delta y}=2y+\lambda(6x+10y)$$
Multiplying the 2 equations by y,x respectively and subtracting I get
$$\lambda(y^2-x^2)=0$$
Hence $$y=x$$
Substituting $x=y$ in $5x^3+6xy+5y^2-8=0$, I get the $x=\pm \frac{1}{\sqrt2}$ and $y=\pm \frac{1}{\sqrt2}$ . Now I am stuck. Both the points corresponds to only one distance. Did I do something wrong?
Best Answer
if the case is $5x^2+6xy+5y^2-8=0$, it could be solved in a easy way:
$5x^2+6xy+5y^2=8,2xy \le x^2+y^2 \implies 5x^2+5y^2 +3(x^2+y^2) \ge 8 \implies x^2+y^2 \ge 1$ when $x=y$
$2xy \ge -(x^2+y^2) \implies 2(x^2+y^2) \le 8 \implies x^2+y^2 \le 4$ when $x=-y$