MHT CET · Maths · Hyperbola
If the tangent at the point \((2 \sec \theta, 3 \tan \theta)\) to the hyperbola \(\frac{x^2}{4}-\frac{y^2}{9}=1\) is parallel to \(3 x-y+4=0\), then the value of \(\theta\) is
- A \(45^{\circ}\)
- B \(60^{\circ}\)
- C \(30^{\circ}\)
- D \(90^{\circ}\)
Answer & Solution
Correct Answer
(C) \(30^{\circ}\)
Step-by-step Solution
Detailed explanation
\(\frac{d}{dx}\left(\frac{x^2}{4}-\frac{y^2}{9}=1\right) \implies \frac{2x}{4} - \frac{2y}{9}\frac{dy}{dx} = 0\) \(m_t = \frac{dy}{dx} = \frac{9x}{4y}\)
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