TS EAMCET · Maths · Hyperbola
If the tangent drawn at the point \(\mathrm{P}(3 \sqrt{2}, 4)\) on the hyperbola \(\frac{x^2}{9}-\frac{y^2}{16}=1\) meets its directrix at \(Q(\alpha, \beta)\) in fourth quadrant then \(\beta=\)
- A \(\frac{5 \sqrt{2}-9}{4}\)
- B \(-\frac{9}{5}\)
- C \(\frac{12 \sqrt{2}-20}{5}\)
- D \(-\frac{5}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{12 \sqrt{2}-20}{5}\)
Step-by-step Solution
Detailed explanation
Tangent equation at \((3 \sqrt{2}, 4)\): \(\frac{x(3 \sqrt{2})}{9} - \frac{y(4)}{16} = 1\) \(\frac{x \sqrt{2}}{3} - \frac{y}{4} = 1\) Eccentricity: \(e = \sqrt{1+\frac{16}{9}} = \frac{5}{3}\) Directrix (for 4th quadrant intersection):…
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