Question

The locus of the point of intersection of two tangents to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ which are inclined at angles θ₁ and θ₂ with major axis such that θ₁ + θ₂ is constant α is

• A. 2xy cot α = x² - y² + b² - a²
• B. 2xy cot α = x² - y² + b² - a²
• C. 2xy cot α = x² + y² + b² - a²
• D. None of these

Answer 1

Answer: (A)

2xy cot α = x² - y² + b² - a²

Answer 2

Proceeding as in Q. 186, we get

m₁ + m₂ = $\frac{2x_{1}y_{1}}{x_{1}^{2} - a^{2}}$ and m₁m₂ = $\frac{y_{1}^{2} - b^{2}}{x_{1}^{2} - a^{2}}$.

Given: θ₁ + θ₂ = constant = α (say)

∴ tan (θ₁ + θ₂) = tan α

⇒$\frac{\tan\theta_{1} + \tan\theta_{2}}{1 - \tan\theta_{1}.\tan\theta_{2}} = \tan\alpha or\frac{m_{1} + m_{2}}{1 - m_{1}m_{2}} = \tan\alpha$

⇒ $\frac{2x_{1}y_{1}/(x_{1}^{2} - a^{2})}{1 - (y_{1}^{2} - b^{2})/(x_{1}^{2} - a^{2})}\tan\alpha$ or

$\frac{2x_{1}y_{1}}{x_{1}^{2} - y_{1}^{2} + b^{2} - a^{2}} = \tan\alpha$

⇒ $2x_{1}y_{1}\cot\alpha = x_{1}^{2} - y_{1}^{2} + b^{2} - a^{2}$.

Hence, locus of P is 2xy cot α = x² - y² + b² - a².