JEE Quadratic #4

Beautiful question! But i took more than prescribed time to solve it.

The given expression can be rewritten as:

$(x-4)^2+(y^2-3)^2=1$

Compare this equation with the equation of unit circle centered at $4,3$ :

$(X-4)^2+(Y-3)^2=1$

$\Rightarrow x=X,y^2=Y$

The expression whose min. and max we need to find reduces to: $X^2+Y^2$ , which is square of the distance of a point on unit circle from origin.

Since center is at $5$ units distance from origin, it's easy to see that minimum distance is $4$ and maximum distance is $6$ .

Hence: $M=36,m=16\Rightarrow M-2m=\boxed{4}$

Raghav's solution is amazing. Here's my not-very-elegant-and-also-cheating-because-I-used-Calculus solution:

First let's use implicit differentiation with respect to $x$ on the original equation:

$\frac { d }{ dx } \left( { y }^{ 4 }-6{ y }^{ 2 }+{ x }^{ 2 }-8x+24=0 \right)$

$4{ y }^{ 3 }\frac { dy }{ dx } -12{ y }\frac { dy }{ dx } +2x-8=0$

Solving for $\frac { dy }{ dx }$ yields:

$\frac { dy }{ dx } =\frac { 4-x }{ 2{ y }^{ 3 }-6y }$

Now let's solve for $x^2+y^4$ in the original equation, and we'll call it $z$ :

$z={ x }^{ 2 }+{ y }^{ 4 }=6{ y }^{ 2 }+8x-24$

Now let's use implicit differentiation with respect to $x$ , substituting for $\frac { dy }{ dx }$ as we go:

$\frac { d }{ dx } \left( z=6{ y }^{ 2 }+8x-24 \right)$

$\frac { dz }{ dx } =12{ y\frac { dy }{ dx } }+8$

$\frac { dz }{ dx } =\frac { 6\left( 4-x \right) }{ { y }^{ 2 }-3 } +8$

Setting $\frac { dz }{ dx }$ to 0 will give us an equation that represents a curve that intersects the curve of the original equation at $z$ 's local extrema. We then solve this equation for $x$ .

$0=\frac { 6\left( 4-x \right) }{ { y }^{ 2 }-3 } +8$

$x=\frac { 4 }{ 3 } { y }^{ 2 }$

We substitute this value for $x$ into the original equation and simplify:

${ y }^{ 4 }-6{ y }^{ 2 }+\frac { 16 }{ 9 } { y }^{ 4 }-\frac { 32 }{ 3 } { y }^{ 2 }+24=0$

$25{ y }^{ 4 }-150{ y }^{ 2 }+216=0$

This yields two solutions: ${ y }^{ 2 }=3.6$ and ${ y }^{ 2 }=2.4$ . The corresponding $x$ values are $x=4.8$ and $x=3.2$ , respectively. Substituting these values into ${ x }^{ 2 }+{ y }^{ 4 }$ yields the maximum $36$ and minimum $16$ .

Needless to say, this took me more than 6 minutes.

(x-4)^2+(y^2-3)^2=1,now take x-4=sin t and y^2-3=cos t,now just putting in the expression we get y^4+x^2=26+6 sin t+8 cos t,so M=36 and m=16,as max 6 sin t+8 cos t is 10 and min is -10, Time Taken around 2 mins

USING CAUCHY SCHWARZ INEQUALITY

From given, we get $x^2 + y^4 = 6x^2 + 8x -24$

Applying cauchy schwarz inequality on the set of real numbers $(6,8),(y^2,x)$ we get
$(6y^2 + 8x)^2 \leq 100(x^2 + y^4)$

Now put $T= x^2 + y^4$ and we need to find out maximum and min values of T

Then the last inequality transforms to
$(T+24)^2 \leq 100T$

$T^2 - 52 T + 576 \leq 0$

$(T-16)(T-36)\leq 0$

$T \in [16,36]$ .
$m=16, M=36$

$M-2m = 4$