सिद्धान्तशिरोमणि: गणिताध्याय (भास्कराचार्य - ग्रहगणित, मध्यमाधिकार व स्पष्टाधिकार सटीक)

268 Fig. 48 But sec z = √(1 + tan² z) = √(x² + y² + 1) ∴ √(x² + y² + 1) / A = y + tan Φ which reduces to x² + y² (1 - A²) - 2A² y tan Φ + 1 - A² tan² Φ = 0 (3) From this it is evident that the locus is an ellipse or parabola or hyperbola according as A ⋚ 1; also it will be seen that the eccentricity is A. The locus is wrongly stated to be always a hyperbola in some text books. For it to be an ellipse A < 1 ie. cos Φ < sin δ ie. δ > 90 - Φ ie. Φ + δ > 90. In such latitudes and under such decli- nations, it will be an ellipse ie. at a place just north of the place where the perpetual day just begins the locus will be an ellipse. Hence in the arctic region it will be always an ellipse; and in the place just at which the perpetual day begins it will be a parabola and in the lower latitudes it will be a hyperbola, ie. it will be a parabola where the latitude Φ is given by 90 - δ. When Φ = 90°, A = cos Φ / sin δ = 0 provided δ ≠ 0. If, however, in addition δ = 0, A becomes indeterminate, but we may note then, that the Sun will be circling round the horizon on that equinoctial day at the north pole. We