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