Let a circle of radius 4 be concentric to the ellipse \( 15x^2 + 19y^2 = 285 \). Then the common tangents are inclined to the minor axis of the ellipse at the angle:
- \(\frac{\Pi}{3}\)
- \(\frac{\Pi}{2}\)
- \(\frac{\Pi}{6}\)
- \(\frac{\Pi}{4}\)
The Correct Option is A
Solution and Explanation
Step 1: Equation of the ellipse
The equation of the ellipse is: \[ \frac{x^2}{19} + \frac{y^2}{15} = 1 \] The minor axis of the ellipse is along the \( y \)-axis.
Step 2: Equation of the tangent line
Let the equation of the tangent line to the ellipse be: \[ y = mx \pm \sqrt{19m^2 + 15} \] We want the common tangents to be inclined to the minor axis at an angle. The equation for the common tangents, when they are parallel to the minor axis of the ellipse, can be written as: \[ mx - y \pm \sqrt{19m^2 + 15} = 0 \] For tangents parallel from the origin \( (0, 0) \) to the circle of radius 4, we get: \[ \frac{\pm \sqrt{19m^2 + 15}}{\sqrt{m^2 + 1}} = 4 \] \[ \Rightarrow 19m^2 + 15 = 16m^2 + 16 \] Simplifying, we get: \[ 3m^2 = 1 \quad \Rightarrow m = \pm \frac{1}{\sqrt{3}} \]
Step 3: Angle of inclination with the \( x \)-axis The angle \( \theta \) of the tangent line with the \( x \)-axis is given by: \[ \theta = \tan^{-1}\left( \frac{1}{\sqrt{3}} \right) \] Thus, \[ \theta = \frac{\pi}{6} \] The required angle is \( \boxed{\frac{\pi}{3}} \).
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Concepts Used:
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