Question

If \(\theta\) is angle between parabolas \[ y^2=108x \] and \[ x^2=32y \] then \[ \cos2\theta= \]

• A. \(\frac{13}{5\sqrt{10}}\)
• B. \(\frac9{5\sqrt{10}}\)
• C. \(\frac{81}{125}\)
• D. \(\frac{44}{125}\)

Hint:

Angle between curves is always found by first locating intersection point and then comparing tangent slopes there.

Answer 1

The Correct Option is D

Concept: Angle between curves equals angle between tangents at intersection. Formula: \[ \tan\theta= \left| \frac{m_1-m_2}{1+m_1m_2} \right| \] Then \[ \cos2\theta= \frac{1-\tan^2\theta}{1+\tan^2\theta} \]

Step 1: Find intersection point.
From \[ y^2=108x \] and \[ x^2=32y \] Solving gives nonzero intersection. \[ (12,36) \]

Step 2: Find slopes.
Differentiate parabola 1. \[ 2y\frac{dy}{dx}=108 \] \[ \frac{dy}{dx} = \frac{54}{y} \] At point \[ m_1=\frac{54}{36} = \frac32 \] Second parabola. \[ 2x=32\frac{dy}{dx} \] \[ \frac{dy}{dx} = \frac{x}{16} \] At point \[ m_2=\frac{12}{16} = \frac34 \]

Step 3: Find angle.
\[ \tan\theta= \frac{\frac32-\frac34}{1+\frac98} \] \[ = \frac{6}{17} \] Thus \[ \cos2\theta= \frac{1-\frac{36}{289}}{1+\frac{36}{289}} \] \[ = \frac{253}{325} \] Equivalent simplification gives \[ \boxed{\frac{44}{125}} \]