AP EAMCET · Maths · Hyperbola
The tangents drawn to the hyperbola \(5 \mathrm{x}^2-9 \mathrm{y}^2=90\) through a variable point P make the angles \(\alpha\) and \(\beta\) with its transverse axis. If \(\alpha, \beta\) are the complementary angles, then the locus of P is
- A \(x^2+y^2=8\)
- B \(x^2-y^2=8\)
- C \(x^2-y^2=28\)
- D \(x^2+y^2=28\)
Answer & Solution
Correct Answer
(C) \(x^2-y^2=28\)
Step-by-step Solution
Detailed explanation
\(\frac{x^2}{18} - \frac{y^2}{10} = 1 \implies a^2=18, b^2=10\) For complementary angles, \(m_1m_2 = 1\). Locus of P is \(x^2-y^2 = a^2+b^2\). \(x^2-y^2 = 18+10\) \(x^2-y^2 = 28\)
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