If $ y^2(y^2-6)+x^2-8x+24=0$ and the minimum value of $x^2+y^4$ is $m$ and maximum value is $M$, then find the value of $m$ and $M$

Question

If $ y^2(y^2-6)+x^2-8x+24=0$ and the minimum value of $x^2+y^4$ is $m$ and maximum value is $M$, then find the value of $m$ and $M$

My Attempt:

I converted them to perfect square i.e., $(x-4)^2+(y^2-3)^2=1$.

Any point on above curve can be written as $x=4+\cos\theta$ and $y^2=3+\sin\theta$

So $x^2+y^4=26+8\cos\theta+6\sin\theta$

As we know $8\cos\theta+6\sin\theta$ varies from $-10$ to $10$

So $x^2+y^4$ must vary from $16$ to $36$.

But given maximum value is $36$ and minimum value is $1$.

can you help me finding minimum value correctly?

Answer 1

Your work seems fine to me.

A geometric approach for the second half (since the bounds on $8 \cos \theta + 6 \sin \theta$ were not obvious to me): Note that if you let $u=y^2$, then you are basically finding the points of the circle defined by $(x-4)^2 + (u-3)^2 = 1$ that are closest/farthest from the origin, which are $\frac{4}{5} (4, 3)$ and $\frac{6}{5}(4, 3)$ respectively, and have squared distances $4^2=16$ and $6^2=36$ from the origin respectively.