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 
 A: 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 $26-10$.
A: Use substitution method:
$$(x-4)^2+(y^2-3)^2=1 \Rightarrow y^2=\pm\sqrt{1-(x-4)^2}+3\\
f(x,y)=x^2+y^4 \Rightarrow \\
f(x)=x^2+1-(x-4)^2\pm6\sqrt{1-(x-4)^2}+9\\
f'(x)=8+\frac{\mp6(x-4)}{\sqrt{1-(x-4)^2}}=0 \Rightarrow \\
4\sqrt{1-(x-4)^2}=\pm3(x-4) \Rightarrow \\
16(-x^2+8x-15)=9(x^2-8x+16) \Rightarrow \\
25x^2-200x+384=0 \Rightarrow \\
x_1=\frac{16}{5}; x_2=\frac{24}{5}\\
f''\left(\frac{16}{5}\right)>0 \Rightarrow f\left(\frac{16}{5}\right)=16 \ \text{(min)}\\
f''\left(\frac{24}{5}\right)<0 \Rightarrow f\left(\frac{24}{5}\right)=36 \ \text{(max)}\\$$
A: By C-S $$0=y^2(y^2-6)+x^2-8x+24=$$
$$=x^2+y^4-8x-6y^2+24\geq$$
$$\geq x^2+y^4-\sqrt{(8^2+6^2)(x^2+y^4)}+24=$$
$$=x^2+y^4-10\sqrt{x^2+y^4}+25-1=\left(\sqrt{x^2+y^4}-5\right)^2-1,$$
which gives
$$4\leq\sqrt{x^2+y^4}\leq6$$ and
$$16\leq x^2+y^4\leq36.$$
The equality occurs for $(|x|,y^2)||(8,6),$ which says that $36$ is a maximal value and $16$ is a minimal value.
