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The curve is the equation of the ellipse $\dfrac{(x-1)^2}{\left(\frac52\right)^2}+\dfrac{(y-2)^2}{\left(\frac53\right)^2}=1,$ and hence $\frac{-3}{2}\leq x\leq \frac72$ . Note that $x^2-2x+y^2-4y+5=(x-1)^2+(y-2)^2=\dfrac{25+5(x-1)^2}{9}.$ The minimum occurs at $x=1$ since $(x-1)^2\geq 0$ , making $\text{min}(x^2-2x+y^2-4y+5)=\frac{25}{9}$ . Also, due to the symmetry of the domain $\frac{-3}{2}\leq x\leq \frac72$ about $x=1$ , substituting the endpoints yields $\dfrac{25+5\left(\frac{-3}{2}-1\right)^2}{9}=\dfrac{25+5\left(\frac72-1\right)^2}{9}=\frac{25}{4},$ making $\text{max}(x^2-2x+y^2-4y+5)=\frac{25}{4}$ . Then $\text{min}(x^2-2x+y^2-4y+5)+\text{max}(x^2-2x+y^2-4y+5)=\dfrac{25}{9}+\dfrac{25}{4}=\dfrac{325}{36}\approx \boxed{9.0278}.$

The first equation represents an ellipse with center as( 1,2) and semi major axis length(a) =2.5 and semi minor axis length(b)=5/3. Now we look at the second equation of which max and min are to be found . The second equation represents a point (1,2). NOw the coordinates of the max an min are (3.5,2 )and (1,1/3) . There are total for possible combination of the max and min coordinates of which one is given above.On substitution of these coordinates into the second equation we get 9.027

The first equation is equivalent to $4 (x-1)^{2}+9 (y-2)^{2}=25$ .

The expression $x^{2}-2x+y^{2}-4y+5=(x-1)^{2}+(y-2)^{2}$

Now we make the substitution $a=(x-1)^{2}, b=(y-2)^{2}$

We have $4a+9b=25$ and we want to find the maximum and minimum of $a+b$ . Note that $a, b \geq 0$ .

Note $4a+4b \leq 4a+9b=25 \implies a+b \leq 6.25$

and $9a+9b \geq 4a+9b=25 \implies a+b \geq \frac{25}{9}=2.777...$

In both cases, equality can be achieved: $(x, y)=(\frac {7}{2}, 2), (1, \frac {11}{3})$ for the first and second case respectively. (There are other equality cases possible but we just need to show one of each.)

Thus the answer is $6.25+2.777...=9.027...$ .

@Mehul Chaturvedi Please see point 2 of Suggestions for Sharers . Only mathematical expressions should be coded in LaTeX.