Question:

The angle between the tangents drawn at \((0,0)\) to the curves \[ y^3-x^2y+5y-2x=0 \] and \[ x^4-x^3y^2+5x+2y=0 \] is

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If the product of slopes of two tangents is \(-1\), then the tangents are perpendicular and the angle between them is \(\frac{\pi}{2}\).
Updated On: Jun 26, 2026
  • \(\frac{\pi}{6}\)
  • \(\frac{\pi}{4}\)
  • \(\frac{\pi}{3}\)
  • \(\frac{\pi}{2}\)
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The Correct Option is D

Solution and Explanation

Step 1: Find the tangent to the first curve at \((0,0)\).
The first curve is \[ y^3-x^2y+5y-2x=0 \] Differentiate implicitly: \[ 3y^2\frac{dy}{dx}-(2xy+x^2\frac{dy}{dx})+5\frac{dy}{dx}-2=0 \] Collecting terms, \[ \left(3y^2-x^2+5\right)\frac{dy}{dx}=2xy+2 \] Thus, \[ \frac{dy}{dx}= \frac{2xy+2}{3y^2-x^2+5} \] At \[ (0,0), \] we get \[ m_1=\frac{2}{5} \]

Step 2: Find the tangent to the second curve at \((0,0)\).
The second curve is \[ x^4-x^3y^2+5x+2y=0 \] Differentiate implicitly: \[ 4x^3-\left(3x^2y^2+2x^3y\frac{dy}{dx}\right)+5+2\frac{dy}{dx}=0 \] Collecting derivative terms, \[ \left(2-2x^3y\right)\frac{dy}{dx} = -4x^3+3x^2y^2-5 \] Thus, \[ \frac{dy}{dx} = \frac{-4x^3+3x^2y^2-5}{2-2x^3y} \] At \[ (0,0), \] we get \[ m_2=-\frac{5}{2} \]

Step 3: Find the angle between the tangents.
For two lines with slopes \(m_1\) and \(m_2\), \[ \tan\theta= \left| \frac{m_2-m_1}{1+m_1m_2} \right| \] Now, \[ m_1m_2= \frac{2}{5}\times \left(-\frac{5}{2}\right) =-1 \] Since \[ 1+m_1m_2=0, \] we get \[ \tan\theta=\infty \] Hence, \[ \theta=\frac{\pi}{2} \]

Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{\pi}{2}} \]
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