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    JEE Main 2026 Exam Analysis April 2, 4, 5 (Out) – Shift Wise Difficulty Level, Paper Review

    Angle of Intersection between Two Curves - Practice Questions & MCQ

    Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

    Quick Facts

    • Angle of Intersection of Two Curves is considered one of the most asked concept.

    • 19 Questions around this concept.

    Solve by difficulty

    If curves $y=f(x)\; \; and\; \; y=g(x)$  are orthogonal at $(x_{1},y_{1}),$ then

    Concepts Covered - 1

    Angle of Intersection of Two Curves

    The angle of Intersection of Two Curves

    Let $y=f(x)$ and $y=g(x)$ be two curves intersecting at a point $P\left(x_0, y_0\right)$. Then the angle of intersection of two curves is defined as the angle between the tangent to the two curves at the point of intersection.

    Let $\mathrm{PT}_1$ and $\mathrm{PT}_2$ be tangents to the curve $\mathrm{y}=\mathrm{f}(\mathrm{x})$ and $\mathrm{y}=\mathrm{g}(\mathrm{x})$ at their point of intersection.
    Let $\Theta$ be the angle between two tangents $\mathrm{PT}_1$ and $\mathrm{PT}_2, \Theta_1$ and $\Theta_2$ are angles made by tangents $\mathrm{PT}_1$ and $\mathrm{PT}_2$ with the positive direction of x -axis, then

    $
    \begin{aligned}
    & m_1=\tan \theta_1 \\
    &=\left(\frac{d}{d x}(f(x))\right)_{\left(x_0, y_0\right)} \\
    & m_2=\tan \theta_2 \\
    &=\left(\frac{d}{d x}(g(x))\right)_{\left(x_0, y_0\right)}
    \end{aligned}
    $

    from the figure $\theta=\theta_1-\theta_2$

    $
    \begin{aligned}
    & \Rightarrow \tan \theta=\tan \left(\theta_1-\theta_2\right)=\frac{\tan \theta_1-\tan \theta_2}{1+\tan \theta_1 \cdot \tan \theta_2} \\
    & \Rightarrow \tan \theta=\left|\frac{\left(\frac{\mathbf{d}}{\mathbf{d x}}(\mathbf{f}(\mathbf{x}))\right)_{\left(\mathbf{x}_{\mathbf{0}}, \mathbf{y}_{\mathbf{0}}\right)}-\left(\frac{\mathbf{d}}{\mathbf{d x}}(\mathbf{g}(\mathbf{x}))\right)_{\left(\mathbf{x}_{\mathbf{0}}, \mathbf{y}_{\mathbf{o}}\right)}}{\mathbf{1}+\left(\frac{\mathbf{d}}{\mathbf{d x}}(\mathbf{f}(\mathbf{x}))\right)_{\left(\mathbf{x}_{\mathbf{0}}, \mathbf{y}_{\mathbf{o}}\right)} \cdot\left(\frac{\mathrm{d}}{\mathbf{d x}}(\mathbf{g}(\mathbf{x}))\right)_{\left(\mathbf{x}_{\mathbf{0}}, \mathbf{y}_{\mathbf{o}}\right)}}\right|
    \end{aligned}
    $

    Orthogonal Curves

    If the angle of the intersection of two curves is a right angle then two curves are called orthogonal curves.
    In this case, $\tan \theta=90^{\circ}$

    $
    \Rightarrow\left(\frac{d}{d x}(f(x))\right)_{\left(x_0, y_0\right)} \cdot\left(\frac{d}{d x}(g(x))\right)_{\left(x_0, y_0\right)}=-1
    $

    This is also the condition for two curves to be orthogonal.
    Condition for two curves to touch each other

    $
    \left(\frac{d}{d x}(f(x))\right)_{\left(x_0, y_0\right)}=\left(\frac{d}{d x}(g(x))\right)_{\left(x 0, y_0\right)}
    $

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    Angle of Intersection of Two Curves

    Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

    Page No. : 5.6

    Line : 67

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