Question

The curve represented by the equation $\frac{x^2}{\sin \sqrt{2}-\sin \sqrt{3}}+\frac{y^2}{\cos \sqrt{2}-\cos \sqrt{3}}=1$ is

• A. An ellipse with the foci on the x-axis
• B. A hyperbola with the foci on the x-axis
• C. An ellipse with the foci on the y-axis
• D. A hyperbola with the foci on the y-axis

Answer 1

Answer: (C)

Since $\sqrt{2}+\sqrt{3}>\pi$, so $0<\frac{\pi }{2}-\sqrt{2}<\sqrt{3}-\frac{\pi }{2}<\frac{\pi }{2}$
and $\cos \left(\frac{\pi }{2}-\sqrt{2}\right)>\cos \left(\sqrt{3}-\frac{\pi }{2}\right)$, i.e. $\sin \sqrt{2}>\sin \sqrt{3}$.
Since $0<\sqrt{2}<\frac{\pi }{2},\frac{\pi }{2}<\sqrt{3}<\pi$, so $\cos \sqrt{2}>0,\cos \sqrt{3}<0$, and $\cos \sqrt{2}-\cos \sqrt{3}>0$.
Thus the curve represented by the equation is an ellipse.
Since $(\sin \sqrt{2}-\sin \sqrt{3})-(\cos \sqrt{2}-\cos \sqrt{3})=2\sqrt{2}\sin \frac{\sqrt{2}-\sqrt{3}}{2}\sin \left(\frac{\sqrt{2}+\sqrt{3}}{2}+\frac{\pi }{4}\right)$
and $\frac{-\pi }{2}<\frac{\sqrt{2}-\sqrt{3}}{2}<0$
we get $\sin \frac{\sqrt{2}-\sqrt{3}}{2}<0,\frac{\pi }{2}<\frac{\sqrt{2}+\sqrt{3}}{2}<\frac{3\pi }{4},\frac{3\pi }{4}<\frac{\sqrt{2}+\sqrt{3}}{2}+\frac{\pi }{4}<\pi ,\sin \left(\frac{\sqrt{2}+\sqrt{3}}{2}+\frac{\pi }{4}\right)>0$ so the expression $(\ast )$ is less than 0 .
That is $\sin \sqrt{2}-\sin \sqrt{3}<\cos \sqrt{3}-\cos \sqrt{3}$, therefore the curve is an ellipse with foci on the $y$-axis.