maximum and minimum of $x^2+y^4$ for real $x,y$ If $y^2(y^2-6)+x^2-8x+24=0$ then maximum and minimum value of $x^2+y^4$ is 
what i try
$y^4-6y^2+9+x^2-8x+16=1$
$(x-4)^2+(y^2-3)^2=1\cdots (1)$
How i find maximum and minimum of $x^2+y^4$ from $(1)$ relation
help me to solve it please 

You could compare equation (1) with a standard ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ where $(x, y)$ has parametric co-ordinates $(acos\theta, bsin\theta)$.
So in your case $(x, y^2)$ could be represented by $(4+cos\theta, 3+sin\theta)$.
Then $x^2 + y^4 = 25 + 1 + 8cos\theta+6sin\theta$ which has a maximum of $26+10$ and a minimum of $26-10$.
