Question

If the angle between the curves \[ y^2=4x \] and \[ y=ax^2-5 \] at the point \((1,2)\) is \(\alpha\), then \[ (a-2)|\tan\alpha| = ? \]

• A. \(1\)
• B. \(3\)
• C. \(\frac{5}{13}\)
• D. \(\frac{13}{5}\)

Hint:

To find the angle between curves, first find the slopes of their tangents at the point of intersection and then apply the tangent-angle formula.

Answer 1

The Correct Option is D

Concept: The angle between two curves at their point of intersection is the angle between their tangents. If slopes are \(m_1\) and \(m_2\), then \[ \tan\alpha = \left| \frac{m_2-m_1} {1+m_1m_2} \right|. \]

Step 1: Determine the value of \(a\).
Since \((1,2)\) lies on \[ y=ax^2-5, \] substitute \[ x=1,\quad y=2. \] \[ 2=a-5. \] \[ a=7. \]

Step 2: Find slope of the parabola \(y^2=4x\).
Differentiate implicitly: \[ 2y\frac{dy}{dx}=4. \] \[ \frac{dy}{dx} = \frac2y. \] At \((1,2)\), \[ m_1=1. \]

Step 3: Find slope of the second curve.
\[ y=7x^2-5. \] \[ \frac{dy}{dx}=14x. \] At \(x=1\), \[ m_2=14. \]

Step 4: Find \(\tan\alpha\).
\[ \tan\alpha = \left| \frac{14-1}{1+14} \right|. \] \[ = \frac{13}{15}. \]

Step 5: Compute the required expression.
\[ (a-2)|\tan\alpha| = (7-2)\cdot \frac{13}{15}. \] \[ = 5\cdot \frac{13}{15}. \] \[ = \frac{13}{3}. \] Hence \[ \boxed{\frac{13}{3}}. \]