Question

If $\theta$ is the acute angle between the tangents drawn from the point (1,5) to the parabola $y^2 = 9x$ then

• A. $\frac{\pi}{6}<\theta<\frac{\pi}{4}$
• B. $\frac{\pi}{3}<\theta<\frac{\pi}{2}$
• C. $0<\theta<\frac{\pi}{6}$
• D. $\frac{\pi}{4}<\theta<\frac{\pi}{3}$

Hint:

Memorizing the direct formula for the angle between tangents from an external point, $\tan\theta = |\frac{\sqrt{S_1}}{x_1+a}|$, is a significant time-saver in competitive exams. It bypasses the need to form and solve a quadratic equation for the slopes.

Answer 1

The Correct Option is D

Step 1: Equation of Tangent: Parabola \( y^2 = 9x \implies 4a=9 \implies a=9/4 \). Equation of any tangent with slope \( m \) is \( y = mx + \frac{a}{m} \). \[ y = mx + \frac{9}{4m} \]
Step 2: Condition for Tangent Passing Through (1,5): Substitute \( x=1, y=5 \): \[ 5 = m(1) + \frac{9}{4m} \] Multiply by \( 4m \): \[ 20m = 4m^2 + 9 \] \[ 4m^2 - 20m + 9 = 0 \] The roots \( m_1, m_2 \) are the slopes of the tangents.
Step 3: Calculate Angle Between Tangents: From the quadratic equation: Sum of roots \( m_1 + m_2 = \frac{20}{4} = 5 \). Product of roots \( m_1 m_2 = \frac{9}{4} \). Difference of roots \( |m_1 - m_2| = \sqrt{(m_1+m_2)^2 - 4m_1m_2} = \sqrt{5^2 - 4(9/4)} = \sqrt{25 - 9} = \sqrt{16} = 4 \). formula for angle \( \theta \): \[ \tan\theta = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right| = \frac{4}{1 + 9/4} = \frac{4}{13/4} = \frac{16}{13} \] So \( \tan\theta \approx 1.23 \).
Step 4: Determine Range: We know: \( \tan(\pi/4) = 1 \) \( \tan(\pi/3) = \sqrt{3} \approx 1.732 \) Since \( 1<1.23<1.732 \), the angle \( \theta \) lies between \( \pi/4 \) and \( \pi/3 \).