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: Understanding the Concept
This problem asks for the angle between the pair of tangents drawn from an external point to a parabola. We can find this angle using a standard formula that relates the angle to the coordinates of the point and the parameter of the parabola.
Step 2: Key Formula or Approach
The angle $\theta$ between the two tangents drawn from an external point $(x_1, y_1)$ to the parabola $y^2 = 4ax$ is given by the formula: \[ \tan\theta = \left|\frac{\sqrt{y_1^2 - 4ax_1}}{x_1+a}\right| \] Alternatively, one can use the slope form of the tangent $y=mx+a/m$, form a quadratic in $m$, and use the angle formula $\tan\theta = |\frac{m_1-m_2}{1+m_1m_2}|$. The direct formula is faster.
Step 3: Detailed Explanation
1. Identify the parabola's parameter and the external point: The parabola is $y^2 = 9x$. Comparing with $y^2=4ax$, we have $4a=9$, so $a = \frac{9}{4}$. The external point is $(x_1, y_1) = (1,5)$. 2. Apply the angle formula: \[ \tan\theta = \left|\frac{\sqrt{y_1^2 - 4ax_1}}{x_1+a}\right| \] Let's calculate the terms inside the formula. The expression under the square root is the value of the parabola's equation at the point, $S_1 = y_1^2 - 4ax_1$. \[ S_1 = 5^2 - 9(1) = 25 - 9 = 16 \] The denominator is $x_1+a = 1 + \frac{9}{4} = \frac{4+9}{4} = \frac{13}{4}$. Now substitute these values into the formula: \[ \tan\theta = \left|\frac{\sqrt{16}}{13/4}\right| = \frac{4}{13/4} = \frac{16}{13} \] 3. Determine the range of $\theta$: We have $\tan\theta = \frac{16}{13}$. Let's compare this value with the tangents of the boundary angles given in the options. $\frac{16}{13} \approx 1.23$. We know: $\tan\left(\frac{\pi}{4}\right) = 1$. $\tan\left(\frac{\pi}{3}\right) = \sqrt{3} \approx 1.732$. Since $1<\frac{16}{13}<\sqrt{3}$, we have $\tan\left(\frac{\pi}{4}\right)<\tan\theta<\tan\left(\frac{\pi}{3}\right)$. Since $\theta$ is an acute angle, the tangent function is increasing in the first quadrant. Therefore, the inequality for the angles is the same: \[ \frac{\pi}{4}<\theta<\frac{\pi}{3} \] Step 4: Final Answer
The tangent of the angle between the tangents is $\frac{16}{13}$, which lies between $\tan(\pi/4)$ and $\tan(\pi/3)$. Thus, the angle $\theta$ lies between $\pi/4$ and $\pi/3$.