Question:

From the point $(-1, -6)$, two tangents are drawn to $y^2 = 4x$. Then the angle between the two tangents is

Updated On: Apr 15, 2025
  • π/3
  • π/4
  • π/6
  • π/2
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The Correct Option is D

Solution and Explanation

We are given the parabola \( y^2 = 4x \) and the point \( (-1, -6) \), and we need to find the angle between the two tangents drawn from this point to the parabola.

Step 1: Equation of the tangents to the parabola
The equation of the tangents to the parabola \( y^2 = 4x \) from a point \( (x_1, y_1) \) outside the parabola is given by the formula: \[ y y_1 = 2(x + x_1) \] where \( (x_1, y_1) \) is the point from which the tangents are drawn. In this case, the point is \( (-1, -6) \). Substituting \( x_1 = -1 \) and \( y_1 = -6 \) into the equation, we get the equation of the tangents: \[ y(-6) = 2(x - 1) \] Simplifying, we get: \[ -6y = 2(x - 1) \] \[ x = -3y + 2 \] 

Step 2: Angle between the tangents
The angle \( \theta \) between two tangents drawn from a point \( (x_1, y_1) \) to the parabola can be found using the formula: \[ \tan \theta = \frac{2 \sqrt{(y_1^2 - 4x_1)}}{y_1} \] Substituting \( x_1 = -1 \) and \( y_1 = -6 \), we get: \[ \tan \theta = \frac{2 \sqrt{((-6)^2 - 4(-1))}}{-6} \] \[ \tan \theta = \frac{2 \sqrt{(36 + 4)}}{-6} \] \[ \tan \theta = \frac{2 \sqrt{40}}{-6} \] \[ \tan \theta = \frac{2 \times 2 \sqrt{10}}{-6} = \frac{4 \sqrt{10}}{-6} = -\frac{2 \sqrt{10}}{3} \] The negative sign indicates that the angle between the tangents is acute. Thus, the angle between the tangents is \( \frac{\pi}{2} \).

Answer:

\[ \boxed{\frac{\pi}{2}} \]

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