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

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∴ b = KA / R - s. But (H sin a) / b = R / S where S is the
shadow so that
S = bR / (H sin a) = (KA / R - s) R / (H sin a) = √(K² - 12²)
∴ (KA - sR) = H sin a √(K² - 12²)
ie. K² A² + s² R² - 2 AsRK = H sin² a (K² - 12²)
∴ K² (A² - H sin² a) - 2 AsRK = - s² R² - 12² H sin² a
∴ K² (H sin² a - A²) + 2 AsRK = 12² H sin² a + s² R² I
This quadratic is of the form ax² + 2bx = c
ie. x² + 2 (b/a) x = c/a II Here 'c'/a is called Adya and b/a Anya.
The solution of the above equation is given by x = - b/a ±
√(b²/a² + c/a)
ie. - Anya ± √(Anya² + Adya). III
When b = KA / R + s, putting - s in the place of s in I,
K = Anya ± √(Anya² + Adya) IV
Out of the four solutions given by III and IV we have
taking the positive solutions
K = √(Anya² + Adya) ± Anya as stated.
Verse 51. If H sin a < A, then in the northern
hemisphere ie. where δ is north,
± √(Anya² - Adya) + Anya = K.
Comm. We have initially put Anya = H sin² a - A².
If H sin a < A, them to avoid a negative value for the