Question:
The hyperbola \(\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1\) passes through the point \((3\sqrt{5}, 1)\) and the length of its latus rectum is \(4/3\) units. The length of the conjugate axis is:
अतिपरवलय \(\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1\) बिंदु \((3\sqrt{5}, 1)\) से होकर गुजरता है और इसके नाभिलंब की लंबाई \(4/3\) इकाई है। संयुग्मी अक्ष की लंबाई क्या है?
Options:
- 2 units
- 3 units
- 4 units
- 5 units
- 2 इकाइयाँ
- 3 इकाइयाँ
- 4 इकाइयाँ
- 5 इकाइयाँ
Correct Answer: c)
Explanation:
Latus rectum of hyperbola: \(\frac{2b^2}{a} = \frac{4}{3} \implies b^2 = \frac{2a}{3}\).
Since it passes through \((3\sqrt{5}, 1)\):
\[\frac{(3\sqrt{5})^2}{a^2} – \frac{1^2}{b^2} = 1 \implies \frac{45}{a^2} – \frac{3}{2a} = 1\]
Multiplying by \(2a^2\):
\[90 – 3a = 2a^2 \implies 2a^2 + 3a – 90 = 0\]
Factorizing: \((2a + 15)(a – 6) = 0 \implies a = 6\) (since \(a > 0\)).
Then \(b^2 = \frac{2(6)}{3} = 4 \implies b = 2\).
The length of the conjugate axis is \(2b = 2(2) = 4\) units.
Hence, the correct option is (c).
नाभिलंब: \(\frac{2b^2}{a} = \frac{4}{3} \implies b^2 = \frac{2a}{3}\)।
बिंदु \((3\sqrt{5}, 1)\) रखने पर:
\[\frac{45}{a^2} – \frac{3}{2a} = 1 \implies 2a^2 + 3a – 90 = 0 \implies a = 6\]
\(b^2 = \frac{2(6)}{3} = 4 \implies b = 2\)।
संयुग्मी अक्ष की लंबाई \(2b = 4\) इकाई है।
अतः सही विकल्प (c) है।