Question:medium
The latus rectum of the hyperbola \(\frac{(3x - 7)^2}{9} - \frac{(4y + 3)^2}{8} = 1\) is
The latus rectum of the hyperbola \(\frac{(3x - 7)^2}{9} - \frac{(4y + 3)^2}{8} = 1\) is
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Be careful with the denominators. In the term \(\frac{(4y+3)^2}{8}\), the 16 from \((4)^2\) moves to the denominator as \(8/16 = 1/2\). Don't just look at the visible denominator '8'.
Updated On: Jun 25, 2026
- \(\frac{1}{2}\)
- \(\frac{16}{3}\)
- 4
- \(\frac{4}{3}\)
- 1
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The Correct Option is
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