The curve represented by the equation

$$\frac{x^{2}}{\sin \sqrt{2} - \sin \sqrt{3}} + \frac{y^{2}}{\cos \sqrt{2} - \cos \sqrt{3}} = 1$$

The standard ellipse is in the form of $$\ \dfrac{\ X^2}{\left(A\right)^2}+\dfrac{\ Y^2}{\left(B\right)^2}=1$$, where the major axis is known by comparing the values of A and B. For instance, If A>B , then major axis is X-axis and focus lie on the X axis, else if B>A, then major axis is Y axis and focus lie on Y axis.

Now, first lets see if a and b , both positive or not :

a = $$Sin\sqrt{\ 2}-Sin\sqrt{\ 3}$$, b =$$Cos\sqrt{\ 2}-Cos\sqrt{\ 3}$$

$$\sqrt{\ 2}\ =\ 1.414\ \ ,\ \sqrt{\ 3}\ =1.732,\ \ \ \dfrac{\ \pi\ }{2}\ =\ 1.5714\ \ $$

Now, Lets use Sin A - Sin B = 2$$Cos\dfrac{\ A+B}{2}Sin\ \dfrac{\ A-B}{2}$$

$$Sin\sqrt{\ 2}-Sin\sqrt{\ 3}\ =\ 2Cos\dfrac{\ \sqrt{\ 2}+\sqrt{\ 3}}{2}Sin\ \dfrac{\ \sqrt{\ 2}\ -\sqrt{\ 3}}{2}$$

                                         = $$2Cos\left(1.573\right)Sin\left(-0.159\right)$$

Now, as 1.573 > $$\ \frac{\ \pi\ }{2}$$ , this implies Cos(1.573) is negative value & Sin(-0.159) is also negative .. Therefore (-Ve)x(-Ve)  is positive . Hence we can conclude 'a' is positive .. Therefore it can be written in the form of $$A^2$$ ..

Similarly, we can check for b :

b =$$Cos\sqrt{\ 2}-Cos\sqrt{\ 3}$$

Now, lets use$$CosA-CosB=\ -2\ Sin\ \dfrac{\ A+B}{2}Sin\ \dfrac{\ A-B}{2}$$    

$$Cos\sqrt{\ 2}-Cos\sqrt{\ 3}$$ =$$-2Sin\left(1.573\right)Sin\left(-0.159\right)$$ 

As both the values 1.573 and 0.159 are less than $$\pi\ $$ , both sin values are positive , but there are two negative signs , hence again positive number. Therefore b > 0.. So this can also be written in the format of $$B^2$$.

Therefore, we can conclude that its a ellipse. Now lets find the major axis by comparing a & b.

Now lets check for a > b :

  $$Sin\sqrt{\ 2}-Sin\sqrt{\ 3}$$  >  $$Cos\sqrt{\ 2}-Cos\sqrt{\ 3}$$,

 $$Sin\sqrt{\ 2}-Cos\sqrt{\ 2}$$  >  $$Sin\sqrt{\ 3}-Cos\sqrt{\ 3}$$,

now, squaring on both sides : { $$Sin^2\left(\theta\ \right)+Cos^2\left(\theta\ \right)\ =\ 1$$,    $$\left(Sin(2\theta\right)\ =\ 2Sin\left(\theta\ \right)Cos\left(\theta\ \right)$$ }

1 - 2($$Sin\sqrt{\ 2}Cos\sqrt{\ 2}$$)  >  1 - 2($$Sin\sqrt{\ 3}Cos\sqrt{\ 3}$$)   

-$$\left(Sin(2\sqrt{\ 2}\right)$$  >  -$$\left(Cos(2\sqrt{\ 3}\right)$$  

This implies ,$$Sin2\sqrt{\ 2}\ \ <\ \ Sin2\sqrt{\ 3}$$

$$2\sqrt{\ 2}\ =\ 2.828\ ,\ which\ is\ less\ than\ \pi\ $$

$$2\sqrt{\ 3}\ =\ 3.464 \ ,\ which\ is\ greater\ than\ \pi\ $$

Hence, we can say that the $$Sin2\sqrt{\ 2}\ >\ 0,\ Sin2\sqrt{\ 3}\ <\ 0$$

Therefore , we can say our assumption of a > b is wrong, and the b > a is true relation.

Hence, we can say Its an ellipse with major axis as Y - axis , therefore the focus lies on Y axis .