a2 + 9b2 +c2 +4d2= 4a + 24b + 6C + 4d - 30
⇒ a2 - 4a + 9b2 - 24b + c2 - 6c +4d2 - 4d +30 = 0
⇒ (a2 - 4a + 4) + (9b2 - 24b +16) + (c2 - 6c + 9) + (4d2 - 4d +1) = 0
⇒ (a - 2)2 + (3b - 4)2 + (c - 3)2 + 4(d2 - d + \(\frac{1}{4}\)) = 0
⇒ (a - 2)2 + (3b - 4)2 + (c - 3)2 + 4(d - \(\frac{1}{2}\))2 = 0 ....(1)
Since,
x2 ≥ 0
Therefore,
x2 + y2 = 0
⇒ x2 = 0 & y2 = 0
Therefore from equation (1), we obtain,
a - 2 = 0, 3b - 4 = 0, c - 3 = 0 & 2(d - \(\frac{1}{2}\)) = 0
⇒ a = 2, b = \(\frac{4}{3}\), c = 3 & d = \(\frac{1}{2}\)
Now,
\(\frac{ad}{bc}\) = \(\frac{2\times \frac{1}{2}}{\frac{4}{3}\times 3}\)
= \(\frac{1}{4}\).
