Question

if α, ß, Γ and δ be the eccentric angles of the feet of four co-normal points of a hyperbola x²/a²-y²/b²=1 from any point in its plane then α+ß+ Γ+ δ can be

Answer 1

Dear student
$$\begin{gathered} \mathrm{Given}:\mathrm{\alpha },\mathrm{\beta },\mathrm{\tau }\mathrm{and}\mathrm{\delta }\mathrm{are}\mathrm{the}\mathrm{eccentric}\mathrm{angles}\mathrm{of}\mathrm{hyperbola} \\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ \mathrm{Then}\mathrm{eccentric}\mathrm{angles}\mathrm{is}\mathrm{given}\mathrm{by}: \\ \frac{\mathrm{x}}{\mathrm{a}}\cos \frac{\mathrm{\alpha }-\mathrm{\beta }}{2}-\frac{\mathrm{y}}{\mathrm{b}}\sin \frac{\mathrm{\alpha }+\mathrm{\beta }}{2}=\cos \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right) \\ \mathrm{and}\mathrm{similarly}, \\ \frac{\mathrm{x}}{\mathrm{a}}\cos \frac{\mathrm{\tau }-\mathrm{\delta }}{2}-\frac{\mathrm{y}}{\mathrm{b}}\sin \frac{\mathrm{\tau }+\mathrm{\delta }}{2}=\cos \left(\frac{\mathrm{\tau }+\mathrm{\delta }}{2}\right) \\ \mathrm{Then}\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\tau }+\mathrm{\delta } \\ =\cos \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)+\cos \left(\frac{\mathrm{\tau }+\mathrm{\delta }}{2}\right) \\ =-2\sin \left(\frac{\left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)+\left(\frac{\mathrm{\tau }+\mathrm{\delta }}{2}\right)}{2}\right)\sin \left(\frac{\left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)-\left(\frac{\mathrm{\tau }+\mathrm{\delta }}{2}\right)}{2}\right)\left[\mathrm{as}\mathrm{cosA}+\mathrm{cosB}=-2\sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\right] \\ =-2\sin \left(\frac{\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\tau }+\mathrm{\delta }}{4}\right)\sin \left(\frac{\mathrm{\alpha }+\mathrm{\beta }-\left(\mathrm{\tau }+\mathrm{\delta }\right)}{4}\right) \end{gathered}$$
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