Question

The curve given by : $\frac{x^2}{\sin \sqrt{2}-\sin \sqrt{3}}+\frac{y^2}{\cos \sqrt{2}-\cos \sqrt{3}}=1$ is?

• A. An Ellipse
• B. A Hyperbola
• C. An open curve.
• D. A closed curve

Answer 1

Answer: (A), (D)

The correct options are
A An Ellipse
D A closed curve

$\sqrt{2}+\sqrt{3}\approx 1.414+1.732=3.146$

$\pi \approx 3.142$

$\Rightarrow \sqrt{2}+\sqrt{3}>\pi$

$\Rightarrow \frac{\pi }{2}-\sqrt{2}<\sqrt{3}-\frac{\pi }{2}$

Now, $\frac{\pi }{2}\approx \frac{3.142}{2}=1.571>1.414=\sqrt{2}$
$\Rightarrow \frac{\pi }{2}-\sqrt{2}>0$

And, $\sqrt{3}=1.732<\pi$

$\Rightarrow \sqrt{3}-\frac{\pi }{2}<\frac{\pi }{2}$
So, $0<\frac{\pi }{2}-\sqrt{2}<\sqrt{3}-\frac{\pi }{2}<\frac{\pi }{2}$
In the interval $(0,\frac{\pi }{2});\cos (...)$ is decreasing function.

Hence,
$\cos (\frac{\pi }{2}-\sqrt{2})>\cos (\sqrt{3}-\frac{\pi }{2})$
ie, $\sin \sqrt{2}>\sin \sqrt{3}$

So, $\sin \sqrt{2}-\sin \sqrt{3}>0$ ... result (i)

Now, $0<\sqrt{2}<\frac{\pi }{2}\Rightarrow cos\left(\sqrt{2}\right)>0$ (In $1^{st}$ Quadrant ${}^{\prime }{\cos }^{\prime }$ is $+ve$)

Also, $\frac{\pi }{2}\approx \frac{3.14}{2}<1.732=\sqrt{3}<3.14=\pi$
ie $\frac{\pi }{2}<\sqrt{3}<\pi \Rightarrow \cos \left(\sqrt{3}\right)<0$ (In $2^{nd}$ quadrant, $cosine$ is $-ve)$

Hence, $\cos \left(\sqrt{2}\right)-\cos \left(\sqrt{3}\right)>0\to result\left(ii\right)$

Hence, we have
$\frac{x^2}{+veTerm}+\frac{y^2}{+veTerm}=1$
Which is clearly an ellipse so option (A)
also; an Ellipse is a closed curve so option (D).