Question:
Find the acute angle at which curves
\[
y=(x-2)^2
\]
and
\[
y=-4+6x-x^2
\]
intersect.
Find the acute angle at which curves
\[
y=(x-2)^2
\]
and
\[
y=-4+6x-x^2
\]
intersect.
Show Hint
To find the angle of intersection of two curves, first find the intersection point and then use the slopes of the tangents at that point.
Updated On: Jun 11, 2026
- \(\tan^{-1}\frac{5}{7}\)
- \(\tan^{-1}\frac{6}{7}\)
- \(\tan^{-1}\frac{4}{7}\)
- \(\frac{\pi}{4}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
The angle between two curves at their point of intersection is the angle between their tangents.
\[
\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|
\]
where \(m_1\) and \(m_2\) are slopes of the tangents.
Step 1: Find the point of intersection.
\[ (x-2)^2=-4+6x-x^2 \] \[ x^2-4x+4=-4+6x-x^2 \] \[ 2x^2-10x+8=0 \] \[ x^2-5x+4=0 \] \[ (x-1)(x-4)=0 \] Thus \(x=1\) or \(x=4\).
Step 2: Find slopes of the curves.
For \[ y=(x-2)^2, \] \[ \frac{dy}{dx}=2(x-2). \] For \[ y=-4+6x-x^2, \] \[ \frac{dy}{dx}=6-2x. \] At \(x=1\), \[ m_1=-2,\qquad m_2=4. \]
Step 3: Apply angle formula.
\[ \tan\theta = \left| \frac{-2-4}{1+(-2)(4)} \right| = \left| \frac{-6}{-7} \right| = \frac67. \] The acute angle is therefore \[ \theta=\tan^{-1}\frac67. \] At the other intersection point the same acute angle is obtained. \[ \boxed{\theta=\tan^{-1}\frac67} \]
Step 1: Find the point of intersection.
\[ (x-2)^2=-4+6x-x^2 \] \[ x^2-4x+4=-4+6x-x^2 \] \[ 2x^2-10x+8=0 \] \[ x^2-5x+4=0 \] \[ (x-1)(x-4)=0 \] Thus \(x=1\) or \(x=4\).
Step 2: Find slopes of the curves.
For \[ y=(x-2)^2, \] \[ \frac{dy}{dx}=2(x-2). \] For \[ y=-4+6x-x^2, \] \[ \frac{dy}{dx}=6-2x. \] At \(x=1\), \[ m_1=-2,\qquad m_2=4. \]
Step 3: Apply angle formula.
\[ \tan\theta = \left| \frac{-2-4}{1+(-2)(4)} \right| = \left| \frac{-6}{-7} \right| = \frac67. \] The acute angle is therefore \[ \theta=\tan^{-1}\frac67. \] At the other intersection point the same acute angle is obtained. \[ \boxed{\theta=\tan^{-1}\frac67} \]
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