Question:
The vertex of the parabola \(9x^2 – 6x + 36y + 9 = 0\) is:
परवलय \(9x^2 – 6x + 36y + 9 = 0\) का शीर्ष क्या है?
Options:
- (1/3, -2/9)
- (-1/3, -1/9)
- (-1/3, 1/2)
- (1/3, 1/2)
- (1/3, -2/9)
- (-1/3, -1/9)
- (-1/3, 1/2)
- (1/3, 1/2)
Correct Answer: a)
Explanation:
Completing the square for \(x\):
\[9\left(x^2 – \frac{2}{3}x\right) + 36y + 9 = 0\]
\[9\left(x – \frac{1}{3}\right)^2 – 9\left(\frac{1}{9}\right) + 36y + 9 = 0\]
\[9\left(x – \frac{1}{3}\right)^2 – 1 + 36y + 9 = 0 \implies 9\left(x – \frac{1}{3}\right)^2 = -36y – 8 = -36\left(y + \frac{2}{9}\right)\]
\[\left(x – \frac{1}{3}\right)^2 = -4\left(y + \frac{2}{9}\right)\]
The vertex \((h, k)\) is \((1/3, -2/9)\).
Hence, the correct option is (a).
पूर्ण वर्ग बनाने पर:
\[9\left(x – \frac{1}{3}\right)^2 = -36\left(y + \frac{2}{9}\right) \implies \left(x – \frac{1}{3}\right)^2 = -4\left(y + \frac{2}{9}\right)\]
शीर्ष \((h, k) = (1/3, -2/9)\) है।
अतः सही विकल्प (a) है।