Question

The angle between the tangents drawn from a point \( (-3, 2) \) to the ellipse \( 4x^2 + 9y^2 - 36 = 0 \) is:

• A. \(45^\circ\)
• B. \( \tan^{-1}\left(\frac{2}{3} \right) \)
• C. \( \tan^{-1}\left(\frac{3}{2} \right) \)
• D. \(90^\circ\)

Hint:

When a point lies on the auxiliary circle or on the director circle of a conic, the angle between the tangents can be exactly \(90^\circ\). Use geometric symmetry to spot such cases.

Answer 1

The Correct Option is D

Step 1: Identify ellipse standard form 
Given ellipse: \( 4x^2 + 9y^2 = 36 \Rightarrow \frac{x^2}{9} + \frac{y^2}{4} = 1 \) 
Step 2: Use formula for angle between tangents from an external point 
Let the point be \( P = (-3, 2) \). 
For an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the angle \( \theta \) between tangents from point \( (x_1, y_1) \) is given by: \[ \cos\theta = \frac{(x_1^2/a^2 + y_1^2/b^2 - 1)^{-1/2}}{\sqrt{\left(x_1^2/a^4 + y_1^2/b^4\right)}} \] Here, \( a^2 = 9 \), \( b^2 = 4 \), \( x_1 = -3 \), \( y_1 = 2 \) Calculate: \[ \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = \frac{9}{9} + \frac{4}{4} = 1 + 1 = 2 \Rightarrow x_1^2/a^2 + y_1^2/b^2 - 1 = 1 \] Now use geometric intuition: From point \( (-3, 2) \), the tangents drawn to the ellipse are perpendicular because the angle between them is \(90^\circ\). Step 3: Conclude 
The angle between tangents is \( \boxed{90^\circ} \).